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  4. SONLI TENGSIZLIKLAR VA ULARNING XOSSALARI. TENGSIZLIKLARNI ISBOTLASH.

SONLI TENGSIZLIKLAR VA ULARNING XOSSALARI. TENGSIZLIKLARNI ISBOTLASH.

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12.08.2023

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8 -MAVZU . SONLI TENGSIZLIKLAR VA ULARNING XOSSALARI. TENGSIZLIKLARNI ISBOTLASH. Reja. 1. Sonli tengsizliklar 2. Sonli tengsizlik larning asosiy xossalari 3. O’rtacha qiymatlar va ular orasidagi munosabatlar 4. Umumlashgan Koshi tengsizligi. 5. Umumlashgan Yung tengsizligi 6. Gel’der tengsizligi. Kalit so ’zlar : Sonli tengsizlik , o’rtacha qiymatla r, Koshi tengsizligi , Yung tengsizligi , Gel’der tengsizligi 1. Sonli tengsizliklar va ularning asosiy xossalari Sonlarni taqqoslash amaliyotda keng qo’llaniladi. Masalan, iqtisodchi rejada ko’zda tutilgan ko’rsatkichlarni amaldagi ko’rsatkichlar bilan taqqos -laydi, shifokor bemorning haroratini sog’lom kishining harorati bilan taqqos laydi, chilangar yo’nayotgan buyumining o’lchamlarini andaza bilan taqqos -laydi. Bu uchala holda qandaydir sonlar o’zaro taqqoslanadi. Sonlarni taqqos -lash natijasida sonli tengsizliklar hosil bo’ladi. 1.1.1 -Ta’rif. Agar &#3627408462; − &#3627408463; ayirma musbat bo’lsa, u holda &#3627408462; son &#3627408463; sondan katta deyiladi. Agar &#3627408462; − &#3627408463; ayirma manfiy bo’lsa, u holda &#3627408462; son &#3627408463; sondan kichik deyiladi. Agar &#3627408462; son &#3627408463; sondan katta bo’lsa, bu &#3627408462; > &#3627408463; kabi; agar &#3627408462; son &#3627408463; sondan kichik bo’lsa, bu &#3627408462; < &#3627408463; kabi yoziladi. Shunday qilib, &#3627408462; > &#3627408463; tengsizlik &#3627408462; − &#3627408463; ayirma musbat, ya’ni &#3627408462; − &#3627408463; > 0 ekanini bildiradi, &#3627408462; < &#3627408463; tengsizlik esa &#3627408462; − &#3627408463; < 0 ekanini bildiradi. 2. Sonli tengsizlik larning asosiy xossalari 1.1 -m iso l. Agar &#3627408475; &#3627408474; to’g’ri kasr bo’lsa, u holda &#3627408475; &#3627408474; < &#3627408475;+1 &#3627408474;+1 bo’lishini isbotlang. ∆ &#3627408475; &#3627408474; kasr &#3627408475; < &#3627408474; bo’lganda (n va m – natural sonlar) to’g’ri kasr deb ata lishini eslatib o’tamiz. Ushbu &#3627408475; &#3627408474; − &#3627408475;+1 &#3627408474;+1= &#3627408475;(&#3627408474;+1)−&#3627408474;(&#3627408475;+1) &#3627408474;(&#3627408474;+1) = &#3627408475;−&#3627408474; &#3627408474;(&#3627408474;+1) ayirma noldan kichik , chunki &#3627408475; − &#3627408474; < 0,&#3627408474; > 0,&#3627408474; + 1 > 0. Binobarin, &#3627408475; &#3627408474; < &#3627408475;+1 &#3627408474;+1 .▲ 1. 2-m iso l. Agar &#3627408462; ≠ &#3627408463; bo’lsa, u holda &#3627408462;2+ &#3627408463;2 > 2&#3627408462;&#3627408463; bo’lishini isbotlang. ∆ &#3627408462;2+ &#3627408463;2− 2&#3627408462;&#3627408463; ayirma musbat ekanligini isbotlaymiz. Haqiqatan ham, &#3627408462;2+ &#3627408463;2− 2&#3627408462;&#3627408463; = (&#3627408462; − &#3627408463;)2 > 0, chunki &#3627408462; ≠ &#3627408463; . ▲ 1. 1-teorema . Agar &#3627408462; > &#3627408463; va &#3627408463; > &#3627408464; bo’lsa, u holda &#3627408462; > &#3627408464; bo’ladi. Δ Shartga ko’ra &#3627408462; > &#3627408463; va &#3627408463; > &#3627408464;. Bu &#3627408462; − &#3627408463; > 0 va &#3627408463; − &#3627408464; > 0 ekanini bildiradi. &#3627408462; − &#3627408463; va &#3627408463; − &#3627408464; musbat sonlarni qo’shib, (&#3627408462; − &#3627408463;)+ (&#3627408463; − &#3627408464;)> 0 ni hosil qilamiz, ya’ni &#3627408462; − &#3627408464; > 0. Demak, &#3627408462; > &#3627408464;. ▲ 1.2 -teorema. Agar tengsizlikning ikkala qismiga ayni bir son qo’shilsa, u holda tengsizlik ishorasi o’zgarmaydi. Δ &#3627408514; > &#3627408515; bo’lsin . Bu holda ixtiyoriy &#3627408464; son uchun &#3627408462; + &#3627408464; > &#3627408463; + &#3627408464; teng -sizlikning bajarilishini isbotlash talab qilinadi. Ushbu (&#3627408462; + &#3627408464;)− (&#3627408463; + &#3627408464;)= &#3627408462; + &#3627408464;− &#3627408463; − &#3627408464; = &#3627408462; − &#3627408463;ayirmani qaraymiz. Bu ayirma musbat, chunki masalaning shartiga ko’ra &#3627408462; > &#3627408463;. Demak, &#3627408462; + &#3627408464; > &#3627408463; + &#3627408464;. ▲ 1.1 -Natija. Istalgan qo’shiluvchini tengsizlikning bir qismidan ikkinchi qis -miga shu qo’shiluvchining ishorasini qarama -qarshisiga almashtirgan holda ko’chirish mumkin. Δ &#3627408514; > &#3627408515; + &#3627408516; bo’lsin . Bu tengsizlikning ikkala qismiga – &#3627408464; sonni qo’shib, &#3627408462; − &#3627408464; > &#3627408463; + &#3627408464; − &#3627408464; ni hosil qilamiz, ya’ni &#3627408462; − &#3627408464; > &#3627408463; ▲ 1. 3-teorema. Agar tengsizlikning ikkala qismi ayni bir musbat songa ko’ paytrilsa, u holda tengsizlik ishorasi o’zgarmaydi. Agar tengsizlikning ikkala qismi ayni bir manfiy songa ko’paytrilsa,u holda tengsizlik ishorasi qarama -qarshisiga o’zgaradi. Δ 1) &#3627408514; > &#3627408515; va &#3627408516; > ?????? bo’lsin . &#3627408462;&#3627408464; > &#3627408463;&#3627408464; ekanini isbotlaymiz. Shartga ko’ra &#3627408462; − &#3627408463; > 0 va &#3627408464; > 0. Shuning uchun (&#3627408462; − &#3627408463;)&#3627408464; > 0, ya’ni &#3627408462;&#3627408464; − &#3627408463;&#3627408464; > 0. Demak, &#3627408462;&#3627408464; > &#3627408463;&#3627408464; . 2) &#3627408514; > &#3627408515; va &#3627408516; < ?????? bo’lsin . &#3627408462;&#3627408464; < &#3627408463;&#3627408464; ekanini isbotlaymiz. Shartga ko’ra &#3627408462; − &#3627408463; > 0 va &#3627408464; < 0. Shuning uchun (&#3627408462; − &#3627408463;)&#3627408464; < 0, ya’ni &#3627408462;&#3627408464; − &#3627408463;&#3627408464; < 0. Demak, &#3627408462;&#3627408464; < &#3627408463;&#3627408464; . ▲ 1. 2-Natija. Agar tengsizlikning ikkala qismi ayni bir musbat songa bo’linsa, u holda tengsizlik ishorasi o’zgarmaydi. Agar tengsizlikning ikkala qismi ayni bir manfiy songa bo’linsa,u holda tengsizlik ishorasi qarama -qarshisiga o’zga -radi. 1.4. -teorema. Bir xil ishorali tengsizliklarni qo’shishda xuddi shu ishorali tengsizlik hosil bo’ladi: agar &#3627408462; > &#3627408463; va &#3627408464; > &#3627408465; bo’lsa, u holda &#3627408462; + &#3627408464; > &#3627408463; + &#3627408465; bo’ladi. Δ Shartga ko’ra &#3627408462; − &#3627408463; > 0 va &#3627408464;− &#3627408465; > 0. Ushbu ayirmani qaraymiz: (&#3627408462; + &#3627408464;)− (&#3627408463; + &#3627408465;)= &#3627408462; + &#3627408464; − &#3627408463; − &#3627408465; = (&#3627408462; − &#3627408463;)+ (&#3627408464;− &#3627408465;). Musbat sonlarning yig’indisi musbat bo’lgani uchun (&#3627408462; + &#3627408464;)− (&#3627408463; + &#3627408465;)> 0,ya’ni (&#3627408462; + &#3627408464;)> (&#3627408463; + &#3627408465;). ▲ 1. 5. -teorema. Chap va o’ng qismlari musbat bo’lgan bir xil ishorali teng -sizliklarni ko’paytirish natijasida xuddi shu ishorali tengsizlik hosil bo’ladi: agar &#3627408462; > &#3627408463;,&#3627408464; > &#3627408465; va &#3627408462;,&#3627408463;,&#3627408464;,&#3627408465;– musbat sonlar bo’lsa , u holda &#3627408462;&#3627408464; > &#3627408463;&#3627408465; bo’ladi. Ushbu ayirmani qaraymiz: &#3627408462;&#3627408464; − &#3627408463;&#3627408465; = &#3627408462;&#3627408464; − &#3627408463;&#3627408464; + &#3627408463;&#3627408464; − &#3627408463;&#3627408465; = &#3627408464;(&#3627408462; − &#3627408463;)+ &#3627408463;(&#3627408464;− &#3627408465;) Shartga ko’ra &#3627408462; − &#3627408463; > 0,&#3627408464; − &#3627408465; > 0,&#3627408463; > 0,&#3627408464; > 0. Shuning uchun (&#3627408462; − &#3627408463;)+ &#3627408463;(&#3627408464;− &#3627408465;)> 0, ya’ni &#3627408462;&#3627408464; − &#3627408463;&#3627408465; > 0, bundan &#3627408462;&#3627408464; > &#3627408463;&#3627408465; . ▲ §2. O’rtacha qiymatlar va ular orasidagi munosabatlar. 1. O’rtacha qiymatlar. a ={a 1, a 2 ,…, a n} musbat sonlar ketma-ketligi uchun o’rta arifmetik qiymat A(a)=A n= naaa n + + + ... 21 , o’rta geometrik qiymat G(a)=G n= n n aaa ... 21 , o’rta kvadratik qiymat K(a)= K n= n aaa n 2 2 2 2 1 ... +++ va o’rta garmonik qiymat N (a)=N n= 1 1 2 1 1 ... − −− +++ naaa n larni aniqlaymiz. Xususan x, y musbat sonlar uchun bu o’rta qiymatlar quyidagicha aniqlanadi: A 2= 2yx + ; G 2= xy ; K 2= 2 22yx + ; N 2 = yx xy + 2. 2. O’rta arifmetik va o’rta geometrik qiymatlar haqida Koshi tengsizligi va uning turli isbotlari. Teorema. A n ≥ G n va An = G n tenglik faqat va faqat a1=a 2 =…= a n tenglik bo’lganda o’rinli. 1-Isboti . x≥ 1 da ex - 1 ≥ x ekanligi ma’lum, ex -1 =x tenglik esa faqat x= 1 da bajariladi . Bundan: 1= e0 = exp ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ∑= 1 )( 1 n i i aA a = )1 )( exp( 1 − ∏= n i i aA a ≥ ∏= n i i aA a 1 )( = n aA а G ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ )()( . Demak, An ≥ G n va tenglik esa faqat 1 () i na Aa = , i= 1, 2 ,…, n bulganda bajariladi. Bundan esa a1=a 2 =…= a n = A n ekanligi kelib chiqadi . 5 n n A G ≥ ekanligini isbotlaymiz: 2 n = da 12 12 2 aa aa + ⋅≤ . Bu tengsizlik ixtiyoriy musbat va sonlar uchun o’rinli bo’lgan 1a 2a () 2 12 0 aa − ≥ tengsizlikdan oson hosil qilinadi. Berilgan tengsizlikni ixtiyoriy p ta natural sonlar uchun to’g’ri deb, p+1 ta natural sonlar uchun to’g ’riligini isbotlaymiz. Bu sonlar bo’lsin. 12, , ..., , nn aa a a +1 1 n a + ularning orasida eng kattasi bo’lsin. Ya’ni, . Shuning uchun 11 1 , ... nnaaa ++ ≥ na ≥ 12 1 ... n n aa a a n + + ++ ≥ . Quyidagicha belgilash kiritamiz: 12 121 1 ... ... , 11 nn n nn aa a aa a a nAa AA nn 1 n n n + + + +++ ++++ ⋅ + === + + . bo’lgani uchun 1 n a + ≥ nA 1 nn aA α + = + deb yozish mumkin, bu yyerda 0 α ≥ . U holda 1 11 nn n nA A A nn nA α α + ⋅+ + == ++ + . Bu tenglikni ikkala qismini ( p+1) – darajaga ko’tarib, quyidagini topamiz: () () () () () ()( )() 1 11 1 11 1 1 ... 11 . n nn nn nnn nn n n nn nn nn AA ACA nn AA AA AA αα αα + ++ ++ + + ⎛⎞ =+ = + + ⎜⎟ ++ ⎝⎠ ≥+⋅=⋅+=⋅ n ≥ Farazga ko’ra, ( ) 12 ... n n n A aa a ≥⋅⋅⋅ . Buni e’tiborga olib, () () 1 11 12 ... nn nnn n 1n A Aa aa aa + ++ ≥⋅≥⋅⋅⋅⋅ + . Bundan 1 112 ... n nn 1 n A aa a a + + + ≥⋅⋅⋅⋅ . Tenglik bo’lganda o’rinli bo’ladi. 12 ... n aa a=== 2-isbot . Teoremaning isboti quyidagi tasdiqqa asoslangan: 6 Agar nomanfiy sonlar 12, , ..., n bb b 12 ... 1 n bb b ⋅ ⋅⋅ = tenglikni qanoatlantirsa, u holda . 12 ... n bb b+++≥ n Bu tasdiqni masalani matematik induktsiya usulida isbotlaymiz. 1 n = da masala ravshan. nk = da 12 ... 1 k bb b ⋅ ⋅⋅ = tenglikni qanoatlantiruvchi ixtiyoriy – nomanfiy sonlar uchun 12, , ..., k bb b 12 ... k bb b k + ++ ≥ tengsizlik o’rinli bo’lsin. da 1 nk =+ 12 1 ... 1 k bb b + ⋅ ⋅⋅ = tenglikni qanoatlantiruvchi ixtiyoriy – nomanfiy sonlar uchun 12, , ..., k bb b +1 12 1 ... 1 kk bb b b + ⋅ ⋅⋅ ⋅ ≥ tengsizlikni qanoatlantirishini ko’rsatamiz. Umumiylikka zarar etkazmasdan 1 1 kbb k+ ≤ ≤ deb hisoblaymiz. Unda bo’lgani uchun induktsiya faraziga ko’ra bo’ladi. Endi () 12 1 1 ... 1 kkk bb b b b −+ ⋅⋅⋅ ⋅ ⋅ = 12 1 1 ... kkk bb b bb k −+ +++ +⋅ ≥ 11 1 kk kkbb bb ++ + ≥⋅ − ekanligini isbotlash etarli. Bu () ( ) 1 1 kkbb + +⋅ −≥ 10 tengsizlikka teng kuchli 1 1 kkbb + ≤ ≤ bo’lgani uchun ohirgi tengsizlik o’rinli ekanligi ravshan. 3-isbot . Teoremaning isboti quyidagi ma’lum tasdiqqa asoslangan: x≥ 1 da ex - 1≥ x , shu bilan birga ex -1 =x tenglik esa faqat x= 1 da bajariladi . Bundan: 1= e0 = exp ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ∑= 1 )( 1 n i i aA a = )1 )( exp( 1 − ∏= n i i aA a ≥ ∏= n i i aA a 1 )( = n aA а G ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ )()( . Demak, A(a) ≥ G(a) va tenglik esa faqat 1 )( = aA a i , i= 1, 2 ,…, n bo’lganda bajariladi. Bundan esa a1=a 2 =…= a n = A(a) ekanligi kelib chiqadi . 1-misol . , 0 xy > bo’lsa, 22 1 x yxyx y + +≥ + + tengsizlikni isbotlang. Yechilishi: 7 2222 22 22 1 222 2 11 1 22 x xyy xy xyxy xy ++≥++⇒ + + + ++ = + + + . 22 2 22 2 , 22 1 ,1 22 1 . 22 xy xy y yxyxyx x x ⎧ +≥ ⎪ ⎪ ⎪ ++≥ ⇒++≥++ ⎨ ⎪ ⎪ +≥ ⎪ ⎩ . y 2-misol . 0 x > bo’lsa, 6 12 4 2222 x x x + ≥⋅ tengsizlikni isbotlang. Yechilishi. 1 11 6 12 4 12 4 6 12 4 2222222 2222 x xxxx x + + ≥⋅ ⋅ =⋅ =⋅ =⋅ x 0 . Misollar: 1. Agar ,xy > bo’lsa, 44 88 x y ++≥ xy 0 ni isbotlang. 2. bo’lsa quyidagini isbotlang: 12 345,,,, xx xxx > ( ) 22222 1 23 45 123 45x xxxx xxxxx ++++≥ +++ . 3. ,, 0x yz > bo’lsa, 222x y z xy yz xz ++≥++ ni isbotlang. 4. bo’lsa, ,, 0abc> 3 abc bca ++ ≥ ni isbotlang. 5. bo’lsa, () ,, 0abc> ( )( )( ) 11 16 abcabc a +++ +≥ bc ni isbotlang. 3. O’rta geometrik va o’rta garmonik qiymatlar orasidagi tengsizli\ k. Teorema. G (a) ≥ H(a) ekanligini, jumladan, H(a) = G(a) tenglik faqat va faqat a1=a 2 =…= a n shart bajarilsa to’g’riligini isbotlang. Isboti . Koshi tengsizligidan foydalanib (1-masalaga qarang) foydalanib 8 ( H(a) ) -1= 1 11 2 1 1 1 1 2 1 1 ))((... ...− −−− − −− = ≥ +++ aGaaa n aaa n n n tenglikga ega bo’lamiz. Jumladan, H(a) = G(a) tenglik faqat a 1=a 2 =…= a n da bajariladi. 1-misol . Agar bo’lsa, ,, 0abc> 3 111 3 abc abc ++ ≤ ++ tengsizlikni isbotlang. Yechilishi : () 111 9 abc abc ⎛⎞ ≤++ ++ ⎜⎟ ⎝⎠ , () 3 3 3 33, 111 99. 111 1 3. abc abc abc abc abc abc abc abc ⎧ ++≥ ⎪ ⎛⎞ ⇒++ ++ ≥ = ⎨ ⎜⎟ ++≥⋅ ⎝⎠ ⎪ ⎩ 2-misol . Agar , ,, 0abc> 23 1 ab c = bo’lsa, 111 6 abc + +≥ ni isbotlang. Yechilishi : 23 6 1 23 1 11111 1 66 abc abbccc ab c ++= +++++≥ = . Misollar 1. ,, 0x yz > bo’lsa, u holda quyidagi tengsizlikni isbotlang: () () 2 222 1111 283 9 xyz xyz xyz x y z ++ +≥ ++ + + . 2. Agar va 12, , ..., 0 n xx x> 12 ... 1 n xx x + ++ = bo’lsa, u holda quyidagi tengsizlikni isbotlang: () () () 12 12 12 1 1 ...1 ... 1 ... 1 ... n n nnx xx xx x xx x x x x − ++ + +++ +++ +++ n ≥ . 3. ,, 0xy z> bo’lsa, u holda quyidagi tengsizlikni isbotlang: 9 222 0 xxyy yzzxz xy yz xz −−− + +≥ +++ . 4. Agar bo’lsa, u holda quyidagi tengsizlikni isbotlang: ,, 0abc> 2 ab c bc ac ab + +≥ +++ . 5. Agar bo’lsa, u holda quyidagi tengsizlikni isbotlang: ,,, 0 abcd > 222 2 2 2 ppp p p p abc abcbacca +++ + + + ++≥ + + b. 4. O’rta arifmetik va o’rta kvadratik qiymatlar orasidagi tengsizlik. Teorema. K (a) ≥ A(a) tengsizlik o’rinli ekanligini, jumladan, K(a) = A(a) tenglik faqat a1=a 2 =…= a n holdagina o’rinli bo’lishini isbotlang. Isboti : Koshi tengsizligidan foydalanib (1-masalaga qarang) foydalanib 2 ijaa ≤ 22 ,1 ijaa i j n + ≤< ≤ tengsizlikni hosil qilamiz. Demak, 222 2 12 1 2 1 (...) ...2 nn ijn aa a aa a aa ≤< ≤ +++ =++++ ij ∑ ≤ 22 2 22 12 1 ... ( ) niijn aa a aa ≤< ≤ ≤++++ + j ∑ = n ( . 22 2 12 ... ) n aa a +++ Eslatib o’tamiz, K (a) = A(a) tenglik faqat a 1=a 2 =…= a n o’rinli bo’ladi. 1-misol . H (a) ≥ min {a1, a 2 ,…, a n} va max {a1, a 2 ,…, a n} ≥ K(a) tengsizliklarni isbotlang. Yechimi: Umumiylikni chegaralamagan holda min{a 1, a 2 ,…, a n}= a 1 , max{a 1, a 2 ,…, a n}= a n deb hisoblash mumkin. U holda 10 H(a)= 11 12 ... n n aa a −− − +++ 1≥ 1 11 1 11 1 ... n a aa a −− − = +++ , K(a)= 22 2 22 2 12 ... ... nnn n n aa a aa aa nn +++ +++ ≤= bo’ladi. Izoh 1 . yuqoridagi misollardan max{a 1, a 2 ,…, a n} ≥ K(a) ≥ A(a) ≥ G(a) ≥ H (a) ≥ min{a 1, a 2 ,…, a n} ekanligi kelib chiqadi . 2-misol . () ( )2 222 3 abc abc ++ ≥++ tengsizlikni isbotlang. Yechilishi : ()()() 222222 22 2 222 333 2 22 0. a b c a b c ab bc ac abc abbcac ab bc ca++≥++ + + ⇒ ⇒++≥++ ⇒− +− +− ≥ 3-misol . ()( ) ( ) ( ) 2 22222 6 ababc ababc +++≥+++ 2 . ⇒ −≥ tengsizlikni isbotlang. Yechilishi: () () () ()() 2 22 2 2222 2 222 2 22 2 0 3 ab ab ababab ab abc abc ⎧ +≥+ ⎪ ×⇒ +≥++ ⎨ ++ ≥++ ⎪ ⎩ ()( ) () ( ) 22 22222 6 ababc ababc +++≥+++ . Misollar 1. Agar va bo’lsa, u holda quyidagi tengsizlikni isbotlang: ,, 0 abc > 1 abc ++= () () () 22 3 abc bac cab +− + + − + + − ≥ 2 . 2. Agar bo’lsa, u holda quyidagi tengsizlikni isbotlang: ,, 0 abc > ()()() 333 222 8 4 abc abc ab bc ca a b b c a c ++ + ≥ ++ + + + . 11 3. Agar bo’lsa, u holda quyidagi tengsizlikni isbotlang: ,, abc R ∈ ()()() () 44 4 444 4 7 ab bc ac a b c +++++≥ ++ . 4. Agar ,, 0xy z> va 3 xy z ++= bo’lsa, u holda quyidagi tengsizlikni isbotlang: xy zx y xz yz + +≥++ . 5. Agar va ,, 0 abc > 222 1 abc + += bo’lsa, u holda quyidagi tengsizlikni isbotlang: 333 5 2 bc ac ab aa bb cc + +≥ − −− . §4. Umumlashgan Koshi tengsizligi. Teorema. a 1, a 2 ,…, a n, p1, p 2 ,…, p n – musbat sonlar bo’lsin. 12 12 ... 11 2 2 12 12 ... ... ... n n pp p p pp nn n n ap a p a p aa a pp p + ++ ⎛⎞+++ ≤ ⎜⎟ +++ ⎝⎠ (1) ekanligini isbotlang, tenglik esa faqat a1= a 2 =…= a n da bajariladi. Isboti: s = 11 2 2 12 ... ... nn n ap a p a p pp p+++ +++ belgilash kiritamiz. ex -1 ≥ x ( x≥1) tengsizlikka ko’ra s ( )1ia e − s≥ a i , i=1, 2,…, n. Bu tengsizliklarni barchasini ko’paytirib chiqamiz: 12 12 12 ... 11 2 2 12 12 ... ... ... exp ... . nn npppp pp nn n n pp p ap a p a p aa a s p p p s s +++ +++ + ++ ⎛⎞ ≤− ⎜⎟ ⎝⎠ = + ++ = Tenglik faqat s =a 1=a 2 =…= a n da bajarilishi esa 1-masaladagidek isbotlanadi. 12 12 12 ... 11 2 2 12 12 ... ... ... n n pp p p pp nn n n ap a p a p aa a pp p + ++ ⎛⎞+++ ≤ ⎜⎟ +++ ⎝⎠ Misol. Quyidagi tengsizlikni isbotlang: 8 346 31618 43 8 abc ab c ⎛⎞ ++ ≥ ⎜⎟ ⎝⎠ . Yechilishi: Koshi tengsizligining umumiy holiga ko’ra p ning o’rnida 3 kelyapti. 1) ;, x Rpq Q ∈ ∈ bo’lsa, () 22 2 sin cos pq pq pq pq xx pq + ⋅≤ + ni isbotlang. 2) 12 234 61220 345 12 abc ab c ⎛⎞ ++ ≥ ⎜⎟ ⎝⎠ 3) bo’lsa, ,, 0 abc > 2 222 22 2 222 47 abc abbcac bca abc ++ ⎛ ⎞ + ++ ≥ ⎜ ++ ⎝⎠ ⎟ isbotlang. 4) bo’lsa, ,, 0 abc > 222 31 abc abbcac bca a b c ++ + ++ ≥ + ++ isbotlang. 5) bo’lsa, ,, 0 abc > 222 26 a b b c a c ab bc ac c a b abc +++ ++ 2 + ++ ≥+ ++ isbotlang. §5. Umumlashgan Yung tengsizligi. Teorema. 1212 12 12 ...... nr rr n n na aa aa a rr r ≤+++ (2) 13 tengsizlik urinli, bu yyerda a1, a 2 ,…, a n, r1, r 2, …, r n lar musbat sonlar, jumladan, 12 11 1... 1 n rr r+++ = . Isboti: 5-masaladagi (1) tenglikda ai ni ga , r ir ia i ni esa 1 ir ( i= 1, 2, …, n) ga almashtirib 1212 12 12 ...... nr rr n n na aa aa a rr r ≤+++ ni olamiz . Izoh . n=2 holida esa Yung klassik tengsizligiga ega bo’lamiz: 11 pqab pq ab + ≥ ( a≥ 0 , b ≥ 0) , (3) bu yyerda p, q sonlar 11 1 pq += tenglikni qanoatlantiruvchi musbat sonlar. 1-misol . Agar va ,, 0 abc > ab bc ac abc + += bo’lsa, bcabc abc bca ≤++ a Yechilishi: Shartga ko’ra 111 1 ab bc ac abc abc + += ⇒++= bcabc abc bca ≤++ a tengsizlik Yung tengsizligin ing xususiy holidan kelib chiqadi. 2-misol. Agar bo’lsa, ,, 0 abc > 236 18 12 6 36a b c abc + +≥ ni isbotlang. Yechilishi: tengizlikni ikkala tomonini 36 ga bo’lamiz 236 18 12 6 36 abc a ++≥ bc 236 12 11 1 ; 236 n abc abc rr r ++≥ +++= … 1 bo’lsa, Yung tengsizligi o’rinli. 236 111 1 236 2 3 6 abc abc ++ =⇒ + + ≥ tengsizlik o’rinli. 14 Misollar 1) Agar ,, 0xy z ≥ va 2 xy z + += bo’lsa, 333 333 2 x y y z z x xy yz zx +++ ++≤ ni isbotlang. 2) Agar va ,, 0 abc ≥ 222 3 abc + += bo’lsa, ()() ( ) 222 ab bc ac −−− 1 ≥ ni isbotlang. 3) Agar va ,, 0 abc ≥ 2 abc + += bo’lsa, 222 22 2 3 ab bc ca ab bc ca ++ + + + ≤ ni isbotlang. 4) Agar va ,, 0 abc ≥ 1 abc + += bo’lsa, () () () 22 3 abc bca cab +− + +− + + − ≥ 2 ni isbotlang. 5) Agar bo’lsa, ,, 0 abc ≥ 4 ab bc ac a b c c a b bcacab +++ ⎛⎞ ++ ≥ ++ ⎜⎟ + ++ ⎝⎠ ni isbotlang. §6. Gel’der tengsizligi. Teorema. 11 1 pq += shartni qanoatlantiruvchi barcha musbat p, q sonlar va aj, bj, j = 1, ..., n sonlar uchun 11 11 1 || || nn n pq p ii i i ii i ab a b == = ⎛⎞⎛ ≤ ⎜⎟⎜ ⎝⎠⎝ ∑∑ ∑ q⎞ ⎟ ⎠ q≠ (4) tengsizlik har doim to’g’ri. Isboti. deb faraz qilamiz (aks holda (4) tengsizlik bajarilishi ravshan). Y ung tengsizligini qo’llab 11 || 0, || 0 nn p ii ii ab == ≠ ∑∑ 15 11 1 11 || || n ii i nn pq pq kk kk ab ab = == ⎛⎞⎛ ⎜⎟⎜ ⎝⎠⎝ ∑∑∑ ⎞ ⎟ ⎠ ≤ 11 1 11 || || || || n ii i nn pq pq kk kk ab ab = == ⎛⎞⎛ ⎜⎟⎜ ⎝⎠⎝ ∑∑∑ ⎞ ⎟ ⎠ ≤ ≤ 1 11 || |||| || pq n ii nn p q i kk kk ab paqb = == ∑∑∑ = 11 1 pq + = ga ega bo’lamiz. Bu yyerdan (4) tengsizlik kelib chiqadi. Izoh . Gyol’der tengsizligining p = q= 2 dagi 22 11 nnn ii i i iii ab a b === ≤ 1 ∑ ∑∑ Koshi-Bunyakovskiy-Shvarts tengsizligi deb ataluvchi bir muhim hususiy holini aytib o’tamiz. 1-misol (Minkovskiy tengsizligi). Ixtiyoriy musbat aj, bj ( j = 1,..., n) sonlar va natural p son uchun () ≤ 1/ () pp kk ab + ∑ ( )1/ pp k a ∑ ( )1/ p k b p ∑ (5) tengsizlikni isbotlang Yechilishi. (a k +b k)p = a k (a k +b k)p-1 + b k (a k +b k)p-1 (k=1, 2, …, n ) tengsizlikni qo’shib, ∑ (a k +b k)p = ∑ a k (a k +b k)p-1 + ∑bk (a k +b k)p -1 ni olamiz. (4) tengsizlikka ko’ra ∑ a k (a k +b k)p-1 ≤ ( )1/ pp k a ∑ ( ) (1)1/ () qp q kk ab − + ∑ , ∑ bk (a k +b k)p-1 ≤( )1/ pp k b ∑ ( ) (1)1/ () qp q kk ab − + ∑ larga ega bo’lamiz, bu yyerdan q (p- 1) = p tenglik yordamida (6) tengsizlik kelib chiqadi. 16 Misollar 1) () () ( ) 11 22 22 2 11 2 2 1 2 1 2 ab a b a a b b +≤+⋅+ 2 Gyol’der tengsizligining 2 p q == holiga ko’ra o’rinli. 2) () 55 555 555 22 333 333 11 2 2 3 3 1 2 3 1 2 3 ababab aaa bbb ⎛⎞⎛ ++ ≤++ ⋅++ ⎜⎟⎜ ⎝⎠⎝ ⎞ ⎟ ⎠ Gyol’der tengsizligining 5 3, , 32 np q 5 = == holiga ko’ra o’rinli. 3) Agar va ,, 0 abc ≥ 333 3 abc + += bo’lsa, ni isbotlang. 44 44 4 4 3 ab bc ca ++ ≤ 4) Agar bo’lsa, isbotlang. ,, 0 abc ≥ () 222 212 a b c abc ab bc ac +++ +≥ ++ 5) Agar va ,, 0 abc ≥ 1 abc= bo’lsa, 22 5 33 abc abc 2 + +++ ≥ ni isbotlang. Amaliyot uchun masalalar. 1-masala. Tengsizliklarni isbotlang: !, 3 n n nn⎛⎞ > ⎜⎟ ⎝⎠ N ∈ ; (1) 1 !,: 2 n n nn N + ⎛⎞ <∀∈ ⎜⎟ ⎝⎠ 2 n ≥ ; (2) 2 !, : n nn nNn>∀∈≥ 3 n≥ ; (3) ; (4) 1 !2 , : 3 n nnN − >∀∈ 17 , 2 nn nn ne nN e⎛⎞ ⎛⎞ << ∀∈ ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ . (5) (1) tengsizlikni isbotlaymiz. Induktsiya bazasi . 1 n= da: 1 1 1! 3⎛⎞ > ⎜⎟ ⎝⎠ ga egamiz. Induktsiya bazasi isbotlandi. Induktiv o’tish. da nk= ! 3 k k k⎛⎞ > ⎜⎟ ⎝⎠ tengsizlik to’g’ri deb faraz qilamiz. da tengsizlik bajar ilishini isbotlaymiz: 1 nk =+ (1) (1) (1)! 3 k k k + + ⎛⎞ +> ⎜⎟ ⎝⎠ .(1)! !(1) (1 3 k k kkk k⎛⎞ += +> + ⎜⎟ ⎝⎠ ) ga egamiz. (1) 1 3 k k + + ⎛⎞ ⎜⎟ ⎝⎠ songa ko’paytiramiz va bo’lamiz: 1 (1) (1) 13 (1) 3 (1) 3 k kk kk kkk k ++ + +⋅⋅+ ⎛⎞ = ⎜⎟ +⋅ ⎝⎠ 11 13 1 1 33 (1 ) kk k kkk + + ++ ⎛⎞ ⎛⎞ => ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ + . Bu yyerda quyidagi joriy hisoblashlarni bajaramiz: 2 11(1)1 (1)...(1)1 (1 ) 1 ... 2! ! k k kkkk k k kk k kk −− ⋅⋅ −+ +=++ ⋅++ ⋅ = 11 11 11 2 1 11 1 ... 1 1 ... 1 2! ! k kkkk k << − ⎛⎞ ⎛⎞⎛⎞⎛ ⎞ =++ − + + − − ⋅ ⋅ − < ⎜⎟ ⎜⎟⎜⎟⎜ ⎟ ⎝⎠ ⎝⎠⎝⎠⎝ ⎠ ��� � ���������� ��� ����������� 18 �N 21 21 11 11 23 2 123...2 11 1 11 1 11 ... 11 ... 2! 3! ! 2 22 k k k k − − =< =< ⋅ ⋅⋅⋅ ⋅ =++ + + + <++ + + + < ��� � 21 11 1 1 1 1 3 1 1 ... ... 1 3 (1 ) 3 1. 11 2 222 1( 2 k kk k k k − <++ + + + + + <+ = ⇒ + < ⇒ > −+ 1) Matematik induktsiya printsip iga asoslanib, ixtiyoriy natural son uchun (1) tengsizlik bajariladi deb xulosa qilamiz. n (2) tengsizlikni isbotlaymiz. Induktsiya bazasi . 2 n = da: (2) tengsizlikning chap tomoni: 2! 2 = ; (2) tengsizlikning o’ng tomoni: 22 21 3 9 2, 25 224+ ⎛⎞⎛⎞ === ⎜⎟⎜⎟ ⎝⎠⎝⎠ . 22 demak, ,25 < induktsiya bazasi isbot bo’ldi. Induktiv o’tish. da nk= 1 !, 2 k k k + ⎛⎞ 2 k < ≥ ⎜⎟ ⎝⎠ tengsizlik to’g’ri deb faraz qilamiz. da 1 nk =+ 1 2) (1)! , 3 k k k + + ⎛⎞ +< ≥ ⎜⎟ ⎝⎠ 2 ktengsizlik bajarilishini isbotlash kerak. 1 (1)! !(1) (1) 2 k k kkk k + ⎛⎞ +=⋅+< ⋅+ ⎜⎟ ⎝⎠ = ga egamiz. 1 2 2 k k + + ⎛⎞ ⎜⎟ ⎝⎠ songa ko’paytiramiz va bo’lamiz: 11 11 11 2 (1)(1)2 2 2(1) 22 2( 2) ( 2) kk kk k < kk k kkk k k kk ++ ++ ++ + + ⋅+⋅ + ⋅+ ⎛⎞ ⎛⎞ == ⎜⎟ ⎜⎟ ⋅+ + ⎝⎠ ⎝⎠ 1 1 2( 1) 1 (2) k k k k + + ⋅+ < + tengsizlik bajarilishi ni isbotlaymiz. 19 1 11 2( 1) 2 1 2 (2) 21 1 11 k kk k k k kk 1 k + + ++ ⋅+ ==⋅ + + ⎛⎞ ⎛ ⎞ + ⎜⎟ ⎜ ⎟ ++ ⎝⎠ ⎝ ⎠ 1 2 0 11(1 )11 (1 ) 1 . . .2 112! 1 (1) k k kkk kk k k + > ++⋅⎛⎞ +=++ ⋅++ ⎜⎟ ++ + + ⎝⎠ ������������ ��� ������������ � > . 1 1 11 2 1 (1 ) 2 2 1 12 1 2 k k k kk + + + ⎛⎞ ⇒+ < ⇒⋅ <⋅= ⎜⎟ ++ ⎝⎠ . 11 22 1 22 kk kk + + ++ ⎛⎞ ⎛⎞ <⋅= ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ . Matematik induktsiya printsipiga asoslanib, ixtiyoriy natural son uchun (2) tengsizlik bajariladi deb xulosa qilamiz. 2 n ≥ (3) tengsizlikni isbotlaymiz. Induktsiya bazasi . 3 n = da: (3) tengsizlikning chap tomoni: 3! 6 36 == ; (3) tengsizlikning o’ng tomoni: 3 2 32 = 7 . 36 27 < demak, induktsiya bazasi isbot bo’ldi. Induktiv o’tish. da nk= 2 !, k kk k > 3 ≥ tengsizlik to’g’ri deb faraz qilamiz. da 1 nk =+ 1 2 (1)! , k kkk + +> ≥ 3 tengsizlik bajarilishini isbotlash kerak. 11 1 22 22 11 22 (1) (1) ( 1) ! ! ( 1) ( 1) ( 1) (1) (1) kk kk kk kkk kkkkk k kk ++ + +⋅ + +=⋅+> ⋅+⋅ =+ ⋅ ++ 1 2 2 2 k k ⋅ > ⋅ 20 (6) masalada tengsizlik isbotlangan edi. Bu tengsizlikdan kelib chiqadi: 1 (1), nnnn n + >+ ∀≥ 3 1 22 (1), nn nn n 3 + >+ ∀≥ . U holda nk = da 1 11 22 22 1 22 1 (1)(1)1 (1) (1) (1 ) (1) (1) k kk k kk kk k kk ++ > +⋅+ >+ ⋅ =+ ⋅+ >+ +⋅ �� ��� �� � 1 1 2 2 k k + . Matematik induktsiya printsipiga asoslanib, ixtiyoriy natural son uchun (3) tengsizlik bajariladi deb xulosa qilamiz. 3 n ≥ (4) tengsizlikni isbotlaymiz. Induktsiya bazasi . 3 n = da: (4) tengsizlikning chap tomoni: 3! 6 36 == ; (4) tengsizlikning o’ng tomoni: 3124 − = . demak, induktsiya bazasi isbot bo’ldi. 64> Induktiv o’tish. (4) tengsizlik nk = da bajariladi deb faraz qilamiz: . da 1 !2 , 3 k kk − >∀ ≥ 3 k 1 nk =+ 11 (1)!2 , k k +− + > ≥ tengsizlik bajarilishini isbotlash kerak. �N 1 1 1 ( 1) ! ! ( 1) 2 ( 1) 2 2 , 3 2 kkk k kkk k k − > + +=⋅+> ⋅+=⋅ > ≥ . Matematik induktsiya printsipiga asoslanib, ixtiyoriy natural son uchun (4) tengsizlik bajariladi deb xulosa qilamiz. 3 n ≥ (5) tengsizlikni isbotlaymiz. Induktsiya bazasi . 1 n= da: 11 11 1! 2 e e⎛⎞ ⎛ ⎞ << ⎜⎟ ⎜ ⎟ ⎝⎠ ⎝ ⎠ . Induktsiya bazasi isbot bo’ldi. 21 Induktiv o’tish. nk = da (5) tengsizlik to’g’ri deb faraz qilamiz: ! 2 kk kk ke e⎛⎞ ⎛⎞ << ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ . da 1 nk =+ 11 11 (1)! 2 kk kkke e + + ++ ⎛⎞ ⎛⎞ <+< ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ tengsizlik bajarilishini isbotlash kerak. Bu tengsizlikning chap tomonini isbotlaymiz. 1 1 (1) 1 (1)! !(1)(1) 1 k kk k k k kk e kkk k eek e + + ⎛⎞ +⋅ ⎜⎟ + ⎛⎞ ⎛ ⎞ ⎝⎠ +=⋅+>+⋅ = ⋅ ⎜⎟ ⎜ ⎟ ⎝⎠ ⎝ ⎠+ ⎛⎞ ⎜⎟ ⎝⎠ = 11 1 1 1(1) 1 1 (1) 1 1 kk kk kk e kkkek ekee k k +++ + ++⋅⋅ + + ⎛⎞ ⎛⎞ ⎛⎞ == ⋅ ⎜⎟ ⎜⎟ ⎜⎟ + ⎝⎠ ⎝⎠ ⎝⎠ ⎛⎞ + ⎜⎟ ⎝⎠ �� ��� �� � 1 k e + > . (5) tengsizlikning chap tomonini isbotlaymiz. 11 1 112 (1)! !(1) 22 2 1 2 k kk k k kk k kkke e e k ++ + ⎛⎞ ⎜⎟ ++ ⎛⎞⎛⎞ ⎛⎞ ⎝⎠ +=⋅+< = ⋅ = ⎜⎟⎜⎟ ⎜⎟ ⎝⎠⎝⎠ ⎝⎠ + ⎛⎞ ⎜⎟ ⎝⎠ k × 11 1 1 21 (1) 1 2 kk kk e kk + + < ≤ + ⎛⎞ ⎛⎞ ×⋅ < ⎜⎟ ⎜⎟ ++ ⎝⎠ ⎝⎠ ��� � �� ��� �� � . Matematik induktsiya printsip iga asoslanib, ixtiyoriy natural son uchun (5) tengsizlik bajariladi deb hulosa qilamiz. n Eslatib o’tamiz, (2) tengnsizlik va 1 1 n e n ⎛⎞ + < ⎜⎟ ⎝⎠ tengsizlikdan foydalanib, da 1 n > 22 111 1 2 ! 22 2 2 nn nn n n n nn n n ke e e e n e + ⎛⎞ ⎛⎞ + ⎜⎟ ⎜⎟ + ⎛ ⎞ ⎛⎞ ⎛⎞⎛⎞ ⎝⎠ ⎝⎠ <=⋅⋅ =⋅⋅<⋅ ⎜ ⎟ ⎜⎟ ⎜⎟⎜⎟ ⎝ ⎠ ⎝⎠ ⎝⎠⎝⎠ ⎛⎞ ⋅ ⎜⎟ ⎝⎠ 2 n n . 2-masala. Tengsizliklarni isbotlang: 1 22...2 21, n nта илдизx nN + =+++ < + ∈ ������ ��� �� �� �� � ; (6) 44...43, n nта илдизx nN =+++ < ∈ ������ ��� �� �� �� � . (7) (6) tengsizlikni isbotlaymiz. Induktsiya bazasi . 1 n= da: 22 2221 nx = +<+ + = 2 (2 1) 2 1 =+= + . Induktsiya bazasi isbotlandi. Induktiv o’tish. n da k = 1 22...2 2 k kта илдизx + 1 = +++ < + ������ ��� �� �� �� � tengsizlik to’g’ri deb faraz qilamiz. da tengsizlik bajarilishini isbotlash kerak: 1 nk =+ 1 2 22...2 2 kkта илдизx + + =+++ < + ��������� �� �� �� � 1. 1 12 1 22...2 221 2221 2 k kта илдизx + +<+ =+++ <++<+ += + ��������� �� �� �� � 1 Matematik induktsiya printsip iga asoslanib, ixtiyoriy natural son uchun (6) tengsizlik bajariladi deb xulosa qilamiz. n (7) tengsizlikni isbotlaymiz. Induktsiya bazasi . 1 n= da: 1 4 x =< 9 . Induktsiya bazasi isbotlandi. 23 Induktiv o’tish. n da k = 44...4 k kта илдизx 3 = +++ < ������ ��� �� �� �� � tengsizlik to’g’ri deb faraz qilamiz. da: 1 nk =+ 1 (3) 444...4 43 kkта илдизx + < 3 = ++++ <+< ��������� �� �� �� � . Matematik induktsiya printsip iga asoslanib, ixtiyoriy natural son uchun (7) tengsizlik bajariladi deb xulosa qilamiz. n 3-masala. 5 5 555 ,: !5!6 n n nNn n − ⎛⎞ ≤∀∈ ⎜⎟ ⎝⎠ 6 ≥ , (8) tengsizlikni isbotlang. Induktsiya bazasi . 6 n = da: 65 65 555 6! 5! 6 − ⎛⎞ ≤ ⎜⎟ ⎝⎠ .Induktsiya bazasi isbotlandi. Induktiv o’tish. 5 5 555 , !5!6 k k k k − ⎛⎞ 6 ≤ ≥ ⎜⎟ ⎝⎠ tengsizlik bajariladi deb faraz qilamiz. 15 15 555 (1)!5!6 k k k +− + ⎛⎞ ≤ ⎜⎟ + ⎝⎠ tengsizlik bajarilishini isbotlash kerak. �N �N 5 5 51 15 5 55 6 5! 6 55555555 ( 1)! ! 1 5! 6 6 5! 6 k kk kk kkk − 5 5 − +− + ⎛⎞< ≤ ⎜⎟ ⎝⎠ ⎛⎞ ⎛⎞ ≤⋅≤ = ⎜⎟ ⎜⎟ ++ ⎝⎠ ⎝⎠ . Matematik induktsiya prints ipiga asoslanib ixtiyoriy natural son uchun (8) tengsizlik bajariladi deb xulosa qilamiz. 6 n ≥ 4-masala. Ixtiyoriy natural son uchun n sin sinkx k x ≤ , (9) tengsizlik o’rinli bo’ lishini isbotlang. 24 Induktsiya bazasi. 1 n= da: sin1 1 sin x x ≤ ⋅ . Induktsiya bazasi isbotlandi. Induktiv o’tish. n da k = sin sin kx k x ≤ tengsizlik bajariladi deb faraz qilamiz. sin( 1) ( 1) sin kxk+≤+ x tengsizlik bajarilishini isbotlash kerak. �N 1 1 sin sin( 1) sin cos sin cos sin cos sin cos kx k x kx x x kx kx x x kx ≤ ≤ ≤ +=+≤⋅+ ≤ �� � � �� � � (1)sinkx ≤ + . Matematik induktsiya printsipiga asoslanib, ixtiyoriy natural son uchun (9) tengsizlik bajariladi deb xulosa qilamiz. n 5-masala. Ixtiyoriy natural son uchun n 11 1 ... 1 12 31 nn n +++ ++ + > , (10) tengsizlik o’rinli bo’ lishini isbotlang. 11 1 ... 12 3 kS kk k =+ ++ ++ + 1 deb belgilab olamiz. Induktsiya bazasi . 1 n= da: 1 11 113 ... 1 11 1 2 311 12 S = +++ = > + +⋅+ . Induktsiya bazasi isbotlandi. Induktiv o’tish. n k = da 11 1 ... 1 12 31 kS kk k = +++ > + ++ tengsizlik bajariladi deb fa raz qilamiz. 1 11 1 1 1 1 ... 1 23 31323334 kS kk k k k k + =+++ + + +++ + + + + > tengsizlik bajarilishini isbotlash kerak. 1 11 1 1 1 1 11 ... 23 31323334 11 kS kk k k k k kk + ⎛⎞ =+++++++− ⎜⎟ ++ + + + + ++ ⎝⎠ = 25 110 111 11111... 1 123 313233341 S kk k k k k k k => > =+ + ++ + + + −> ++ + + + + + + ������������� ����������� ������������� ���������� � . 1111112 323334 1323433 kkkkkkk ++ −= + − +++++++ = (3 4)(3 3) (3 2)(3 3) (6 4)(3 4) 2 0 (3 2)(3 3)(3 4) (3 2)(3 3)(3 4) kk kk kk kkk kkk +++++−++ == +++ +++ > " ekanligidan tengsizlik kelib chiqadi. Mate matik induktsiya printsipiga asoslanib, ixtiyoriy natural son uchun (10) tengsizlik ba jariladi deb xulosa qilamiz. "0> n 6-masala. 222x y z xy xz yz ++≥++ tengsizlikni isbotlang, bu yyerda ,,x yz - musbat sonlar. Yechilishi. Ma’lum 2 2 22 22 2, 2, 2 x yxyxzxzyz +≥ +≥ +≥ yz ) tengsizliklarni qo’shib, ushbu 22 22 22 222 ()()()2( )2(x y x z y z x y z xy xz yz +++++≥ ++≥ ++ tengsizlikni olamiz. 7-masala. 444 ( ) x y z xyz x y z ++≥ ++ tengsizlikni isbotlang, bu yerda , ,x yz - musbat sonlar. Yechilishi. 1-masalaga ko’ra: 444 22 22 2222222 () ( ) () 2 x yz x y z xyyzxz ++= + + ≥ + + ga egamiz. Bu yerdan esa 2 2 22 22 ( ) x y y z x z xyyz yzzx zxxy xyz x y z ++≥++= ++ ni olamiz. 8-masala. 4444 4 x y z u xyzu +++≥ tengsizlikni isbotlang, bu yerda , , ,x yzu - musbat sonlar. Yechilishi. 44 2244 2 2, 2 2 x yxyzuz +≥ +≥ u 2 ga egamiz. Demak, 4444 22 2 22 x yzu xy zu +++≥ + . Bundan tashqari 22 22 2 x yzu xyz +≥ u. Demak, 4444 4 x y z u xyzu +++≥ . 26 9-masala. 2 11()() 24 x yxyxyy ++ +≥ + x tengsizlikni isbotlang, bu yerda ,x y- musbat sonlar. Yechilishi. Birinchidan, 2 111 ()()()( 242 xy xy xyxy 1 ) 2 + ++=+ ++ . Ikkinchidan, 2 xy xy + ≥, 111 244 x yxy x ++ =+ ++ ≥ + y. Demak, 2 11 ()() ( ) 24 x yxyxyyxxyy ++ += + ≥ + x. 10-masala. 0, 0 x y ≥≥ va 2 x y + = bo’lsin. 22 2 2 () xy x y +≤ 2 1 tengsizlikni isbotlang. Aniqlik uchun 1, 1,0 x y ε εε = +=−≤≤ deb olamiz. U holda 22 2 2 2 2 2 2 22 2 ( ) (1 )(1 )((1 ) (1 )) (1 )(2 2 ) xy x y εε ε ε ε ε +=− + −++ =− + = 222 24 2(1 )(1 )(1 ) 2(1 )(1 ) 2 εεε εε =−−+=−−≤ 11-masala . va b bir xil ishorali sonlar bo’lsin. a 22 2 2 2 3 () 10 41 2 ab a b a ab b ++ ≤ + ekanligini isbotlang. Qachon tenglik bajariladi? ekanligini hisobga olib va Koshi tengsizligidan foydalanib quyidagiga ega bo’lamiz: 0 ab > 22 22 3 2 () 10 4 23 1 aabb ab ab ab a abb ab ab ++ ++ ++ ⋅⋅ ≤ = 2+ . Tenglik esa bo’lganda bajarilishini eslatib o’tamiz. ab= 12-masala . va birdan katta sonlar bo’lsin. , ab c 22 2 log ( ) log ( ) log ( ) 1 ab b bca bac cab abc ac ab bc −+ −+ − + ≥ tengsizlikni isbotlang. 27 1, 1, 1 abc >>> va 22 2 2, 2, 2 bca ac b ab c bc a ac ab bc +≥ +≥ +≥ bo’lgani uchun 22 2 log ( ) log ( ) log ( ) log (2 ) ab b a bca bac cab abc bb ac ab bc − +−+−+≥ −× = . log (2 ) log (2 ) log log log 1 bc abc cc aa b c a ×− −= 13-masala . va b musbat sonlar bo’lsin. a 33 3 11 2( )( ) ab ab ba ab + ≤++ tengsizlikni isbotlang. Yechilishi. Berilgan tengsizlikni kubga ko’taris h va soddalashtirishlardan so’ng quyidagiga ega bo’lamiz: 33 3( ) 4 ab a ba b +≤++ b a . Koshi tengsizligiga ko’ra 3 11 3 a bb ++ ≥ a va 3 11 3 b aa ++ ≥ b tengsizliklarni olamiz va ularni qo ’shib, qidirilayotgan tengsizlikni olamiz. 14-masala. 2 1 2131 ( 31 4(1) n kn nn nN nkn =+ ) + ≤≤ ∈ ++ ∑ ekanligini isbotlang. Yechilishi. 22 22 11 11 111111131 () 31 2 31 2 (31) nn nn kn kn knk n n knk knk kkn =+ =+ =+= + + ==+=≤ +− +− +− ∑∑ ∑ ∑ k ; 1) 22 2 (3 1) 3 1 (3 1) (3 1 ) ( ) 44 4 nn n kn k k ++ + +− = − − ≤ tengsizlikdan 2 2 1 1314(31)2 2(31)2(31)3 n kn nnn kn k n n =+ ++ ≥= +− + + ∑ 1 n ekanligi kelib chiqadi; 28 2) (3 1 ) 2 ( 1) (2 )( ( 1)) 0 ( 1 2 ) kn k nn nkk n n k n +− − + = − − + ≥ +≤ ≤ tengsizlikdan 2 1 131(31)32 (31)4(1)4(1 n kn nnnn kn k nn n =+ 1 ) + ++ ≤= +− + + ∑ ekanligi kelib chiqadi. Demak 2 1 2131 () 31 4(1) n kn nn nN nkn =+ + ≤≤ ∈ ++ ∑ . 15-masala . va musbat sonlar bo’lsin. tengsizlik va faqat biror uchburchak tashkil qilganda gina bajarilishi mumkinligini isbotlang. , ab c 444 222 2( ) ( ) abc abc ++ < ++ 2 0 = , ab c Yechilishi. Ravshanki, bizning tengsizligimiz 444 22 22 22 222 abc ab bc ac ++− + + < tengsizlikka teng kuchli. Oxirgi tengsizlikning chap tomonini almashtiramiz: 444 22 22 22 222 22 2 22 2 22 22 22 2224 ( 2 )( 2 ) (( ) )(( ) ) ()()()() abc ab bc ac abc ab abc ababc ab ab c ab c a b ca b ca b ca b c++− + + = +−− = =+−− +−+ =−− +− =−+−−+++− Demak, berilgan tengsizlik va biror uchburchak tashkil qilganda aniq ravishda bajariladigan ushbu , ab c ( )( )( )( ) 0 abcabcabcabc − +−−+++−> tengsizlikka teng kuchli. Endi faraz qilamiz, bu tengs izlik bajariladi, biroq va biror uchburchak tashkil qilmaydi. U holda , ab c , , , abcabcabcabc − +−−+++− sonlardan kamida ikkitasi manfiy. va 0 abc +−< 0 bca + −< bo’lsin. Bu yerdan masalaning shartiga zid bo’lgan tengsizlikni olamiz. 2 0b < 16-masala . – ketma-ketlikning biror o’rin almashtirishi bo’lsin. 12,,..., n bb b 12,,..., n aa a 12 12 11 1 ( ) ( )...( ) 2 n n n aa abb b ++ +≥ 29 tengsizlikni isbotlang. Tengsizlikning isboti Koshi tengsizligidan darhol kelib chiqadi: 12 12 12 12 12 12 ... 11 1 ( ) ( ) ...( ) 2 2 ...2 2 ... n nn n nn aaaa aa aa a bb b bbbbbb ++ +≥ = = n . 17-masala. 106532 10 x xxxxx ++++++> tengsizlikni isbotlang. Yechilishi. Ravshanki, tengsizligimiz 0 x≥ da bajariladi, shuning uchun xmanfiy bo’lgan qiymatlarni qarash etarli. 1 x≤ − bo’lsin, holda 10 5 0 x x + ≥ , 63 0 x x +≥ , 2 0 x x +≥ va 1 tengsizliklarni qo’shib, iz lanayotgan tengsizlikni olamiz. 0 > Endi 1 0 x −< < bo’lsin. Qism hollarni qaraymiz: a) 5 10 x x ++> . U holda 106532 1062 3 5 11( xxxxxx xxx x xx ++++++= +++++ ++> 1) 0 . b) 5 10 x x ++≤ . U holda 106532 55 2355 1 ( 1) ( ) (1 ) ( 1) 0 xxxxxx xxx xx xxxx + + + + ++= ++ + + + + > ++ ≥ . 18-masala . 1, 1 x y >− >− va bo’lsin. 1 z >− 222 22 11 1 2 111 xyz S yz zx xy ++ + =++ ++ ++ ++ 2≥ tengsizlikni isbotlang. 22 2 11 11 2 x x yz y z ++ ≥ ++ + + tengsizlik yordamida ,x y va z ning manfiy bo’lmagan qiymatlarini qarash etarli. 222 22 2 22 11 1 () 111 111 zx xy yz z x y S yz zx xy yz zx xy ++ ++ + + =++− ++ ++ ++ ++ ++ ++ ++ 2 ≥ 222 3 22 2 2 2 11 1 3( 111 1 1 1 zx xy yz z x y yz zx xy yz zx xy ++ ++ + + ≥⋅⋅−++ ++ ++ ++ ++ ++ ++ 2)≥ 22 2 3( ). 111 zxy yz zx xy ≥− + + ++ ++ ++ 30 Endi 122 1 111 zxy S yz zx xy =++ ++ ++ ++ 2≤ ekanligini isbotlaymiz. 0 x= holni qaraymiz. U holda. Demak, 0 xyz = da 1 1 S≤ . 0 xyz ≠ holda 1 11 111 11 1 () () ( ) 2 2 2 S zy xzy xz y 1 x x xzz yyxz =++ ≤++ ++ ++ + + + + + y . , xy a yz == b va z c x = deb belgilab olamiz. 1 abc= va 1 111 222 S abc =++ +++ ekanligi ravshan. U holda 1 1 1 (2 )(2 ) (2 )(2 ) (2 )(2 ) 222 (2)(2)(2) bc ac ab abc abc ++++++++ ++= +++ +++ = 12 4( ) ( ) 84( )2( ) abc abbcac abc abbcac abc ++++++ =≤ ++++ +++ 222 3 12 4( ) ( ) 12 4( ) ( ) 1. 12 4( ) ( ) 84( )( )3 a b c ab bc ac a b c ab bc ac abc abbcac a b c ab bc ac a b c abc ++++++ ++++++ ≤= ++++++ +++++++ + = Demak, . 1 1 S ≤ 19-masala . Ixtiyoriy musbat , ( 1, 2,..., )jjab j n = sonlar uchun 11 11... ... ( )...( ) nnn nn naa bb a b a b+≤+ + n tengsizlik o’rinli ekanligini isbotlang. Yechilishi. Gyuygens tengsizligiga asoslanib 11 11 1...1 1 ... n nn n nn aa aa bb bb ⎛⎞ ⎛⎞ ⎛⎞ ++≥+ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ ⎝⎠ yoki ( ) 11 1 1 ( )...( ) ... ... n nn nn n n ab a b aa bb++≥ + ni olamiz. Bu yerdan esa 11 11... ... ( )...( ) nnn nn naa bb a b a b+≤+ + n kelib chiqadi. 31 20-masala . Ixtiyoriy musbat sonlar uchun 12,,..., n aa a 2 12 12 11 1 (...) .. n n aa a n aa a ⎛⎞ +++ + ++ ≥ ⎜ ⎝⎠ ⎟ tengsizlik o’rinli bo’lishini isbotlang. Yechilishi. Koshi–Bunyakovskiy–Shvarts tengsizligiga ko’ra () 2 2 12 12 2 22 22 2 12 2 12 11 1 ... 11 1 ( ) ( ) ... ( ) .. n n n n na a aaa a aa aa aa a ⎛⎞ =⋅+⋅++⋅ ≤ ⎜⎟ ⎜⎟ ⎝⎠ ⎛⎞⎛⎞ ⎛⎞⎛ ⎞ ⎜⎟ ≤+++ ⋅ + ++ ⎜⎟ ⎜⎟⎜ ⎟ ⎜⎟⎜ ⎟ ⎜⎟ ⎜⎟ ⎝⎠⎝ ⎠ ⎝⎠ ⎝⎠ ni olamiz. 21-masala. ( )2 12 12 22 2 12 1334 1 ... ... ... n n n aa a aa a aa a aa aa aa +++ ≤+ ++ +++ + + + 2 n tengsizlikni isbotlang, bu yerda . 0 ( 1, 2,..., ) kak≥= Yechilishi. Koshi–Bunyakovskiy–Shvar ts tengsizligiga ko’ra 2 2 1 12 11312 13 12 ( ... )( ) ...( ) n n n aa aa a aaaaaa aa aa ⎛⎞ +++ = ⋅ + ++ ⋅ + ⎜⎟ ⎜⎟ ++ ⎝⎠ ≤ () 12 11 3 1 2 13 3 4 12 ... ( ) ... ( ) n n aa aaa a a a a aa a a aa ⎛⎞ ≤+++ +++≤ ⎜⎟ ++ + ⎝⎠ 22 22 12 12 13 13 3 4 12 2 2 22 22 221112 11 ... ( ) ( ) ... 22 11 11 ( )()()() 22 22 n nn n n n aa a aa aa aa a a aa a a aa aa aa − ⎛⎞ ⎛⎞ ≥+++ ++++ ⎜⎟ ⎜⎟ ++ + ⎝⎠ ⎝⎠ ⎛⎞ ⎛ + ++++ +++= ⎜⎟ ⎜ ⎝⎠ ⎝ + ⎞ ⎟ ⎠ 22 12 1 13 3 4 12 ... (2 ... 2 ) n n aa aaa aa a a aa ⎛⎞ =+++ +++ ⎜⎟ ++ + ⎝⎠ . 32 Mashqlar 1. Agar , bo’lsa, u holda tengsizlik o’rinli bo’ lishini isbotlang. ,,, 0, , abcd c d ac d b >+≤+≤ ab bc ab+≤ 2. Agar ,,xy z lar xaqiqiy sonlar to’p lamiga tegishli bo’lsa, 222x y z xy yz xz ++≥++ tengsizlikni isbotlang. 3. Agar 1 xy z ++= bo’lsa, 222 1 3 xyz + +≥ ni isbotlang. 4. Agar bo’lsa, 0 ab > 2 ab ba + ≥ tengsizlikni isbotlang. 5. Xar qanday ( 2 n ≥ nN∈ ) larda 22 2 11 1 1. 23 n + ++ <… tengsizlik o’rinli bo’lishini isbotlang. 6. Xar qanday ( n ) larda 2 n ≥ N ∈ 11 1 1 223 2 n n <+ + + + − … 1 tengsizlik o’rinli bo’lishini isbotlang 7. bo’lsa, nN ∈ () 2 11 1 925 21n +++ + … tengsizlikni isbotlang. 8. bo’lsa, nN∈ 11 1 1 212 2 nn n <+ ++< ++ … 3 4 2 tengsizlikni isbotlang. 9. Agar bo’lsa, tengsizlikni isbotlang. 22 222 2 12 12 1 nn aa a bb b+++ =+++= …… 11 2 2 11 nn ab a b a b −≤ + + + ≤ … 10. Agar bo’lsa, tengsizlikni isbotlang. 12 1 2 1, , , 0 nn aa a a a a⋅⋅ = > …… ()( ) () 12 11 1 n n aa a ++ +≥ … 11. Agar bo’lsa, 1 ab +≥ 44 1 8 ab + ≥ tengsizlikni isbotlang. 33 12. musbat sonlar va birdan farqli bo’lsa, , ab log log 2 abba +≥ tengsizlikni isbotlang. 13. 2 11 2 log log 2 π π + > tengsizlikni isbotlang. 14. Agar bo’lsa, nN∈ 11 1 1 1! 2 ! ! n 3 + +++ < … tengsizlikni isbotlang. 15. Agar bo’lsa, nN∈ ( ) 11 1 2111 2 23 nn n +− <+ + + + < … tengsizlikni isbotlang. 16. Agar ( 12 kk k aa a − =+ − 3, 4, . k = … ) bo’lsa, 2345 11 2 3 5 2 2 2222 2 n na ++++++< … tengsizlikni isbotlang. 17. Agar n bo’lsa, N ∈ 11 1 12 12 31 nn n < +++ < + ++ … tengsizlikni isbotlang. 18. Agar bo’lsa, 0, 1, 2, , iai>= … n 12 12 n n naa a aa a n + ++ ≥⋅⋅… … tengsizlikni isbotlang. 19. Agar bo’lsa, 0, 1, 2, , iai>= … n () 2 12 12 11 1 n n aa a n aa a ⎛ +++ + ++ ≥ ⎜ ⎝⎠ …… ⎞ ⎟ tengsizlikni isbotlang. 20. Agar bo’lsa, 0 a > 22 221 1 n n aa a n n aa a − 1 + +++ + ≥ + ++ … … tengsizlikni isbotlang. 34 21. Agar 12 02 n π αα α << << < … bo’lsa, 12 1 12 sin sin sin cos cos cos n n n tg tg α αα α α αα α +++ < +++ … … < tengsizlikni isbotlang. 22. Agar n bo’lsa, quyidagi tengsizlikni isbotlang. N ∈ () ( ) () 2! 4! 6! 2 ! 1 ! n nn ⋅⋅⋅ ⋅ ≥ + … 23. Agar 0 2 π ϕ << bo’lsa, 1 2 ctg ctg ϕ ϕ ≥+ tengsizlikni isbotlang. 24. butun sonlar va , kl − 2n α βπ ≠ ±+ bo’lsa, 22 cos cos cos cos cos cos kl lk kl αβ α β αβ − ≤ − − tengsizlikni isbotlang. 25. Agar bo’lsa, () tengsizlikni isbotlang. 2 n > 2! n n > n 26. Agar va ,, , 0 ab pq > ,p q ratsional sonlar 11 1 pq + = shartni qanoatlantirsa pqab ab p q ≤+ tengsizlikni isbotlang. 27. Agar n bo’lsa, N ∈ 1 21 n n ⎛⎞ 3 < +< ⎜⎟ ⎝⎠ tengsizlikni isbotlang. 28. Agar bo’lsa, quyidagi tengsizlikni isbotlang. 0 n > ! 3 n n n ⎛⎞ < ⎜⎟ ⎝⎠ 29.Agar bo’lsa, quyidagi tengsizlikni isbotlang. () 0 n > () 3 3! n nn > 30. Agar 12 ;0,1,2,ni , s aa aa i =+ ++ > = …… n bo’lsa, ()( ) () 2 12 11 1 1 1! 2 ! ! n n s s aa a n ++⋅⋅+≤++++ … s … tengsizlikni isbotlang. 35 31. a) 2 13 1 a aa ≤ ++ ; b) 2 1 2 49 a aa ≤ − + ; s) 2 2 1 aa a a +≥+ 1 ; d) 4 2 16 2 4 a a a + ≥ + . 32. a) 2 93025aa −+≥ 0; b) 2 25 10 bb +≥ ; s) 2 452 2 aa a −+≥ − ; d) 2 2106 bb b 1 − +≥ − . 33. a) ; b) 44 3ab abab+≥ + 3 0 44 22 () ababab + ++ ≥ 5 ; s) ; d) 6 6 42 24a b ab ab+≥ + 66 5ab abab + ≥+ . 34. Agar va bo’lsa, u holda quyidagilarni isbotlang: 0 a ≥ 0 b ≥ a) ; b) ; 33 2ab abab+≥ + 2 3 ) 4 3 33 ()4(ab a b +≤ + s) ; d) 55 4ab abab+≥ + 55 32 2ab abab + ≥+ . 35. Agar va bo’lsa, u holda quyidagilarni isbotlang: 0 a ≥ 0 b ≥ a) ; b) 33 2ab abab−≥ − 2 ) 33 3( ab abab − ≥− ; s) ; d) 33 2 2ab ab a−≥ − b 4 55 4ab abab − ≥− . 36. va sonlarning ixtiyoriy qiyma tlarida tengsizlik o’rinli a b bo’lishini isbotlang: a) ; 43 22 34 22 2 aabababb −+ − +≥ 0 2) 2) b) . 43 22 3 4 4 8 16 16 0 aabab ab b −+ − +≥ 37. Ixtiyoriy , va sonlar uchun tengsizlik ,, abc d a) ; 222 2 ()()(a b c d ac bd −−≤− b) . 222 2 ()()(a b c d ac bd ++≥+ o’rinli bo’lishini isbotlang, jumladan tenglik bajariladi shu holda va faqat shu holdaki, qachonki ad bc = . 36 38. shartni qanoatlantiruvchi ixtiyoriy a va b sonlar uchun tengsizlik o’rinli bo’lishini isbotlang. 0 ab ≥ 222 ()(ab ab −≥− 4) 39. Agar bo’lsa, ab< 2 ab a b + << bo’lishini isbotlang. 40. Agar bo’lsa, abc<< 3 abc a b + + < < bo’lishini isbotlang. 41. Agar ekanligi ma’lum. 0, 0, 0, 0 abcd >>< < ,,,,,, ,, ab ac c b ac abd abc bcd abcd c d ad cd bd c ifodalar qanday ishoralarga ega bo’ladi ? 42. Agar a) va b bir xil ishorali sonlar; a b) va b turli ishorali sonlar ekanligi ma’lum bo’lsa, a ab ko’paytma va a b kasr qanday ishoralarga ega bo’ladi ? 43. Agar a) ; b) 0 ab > 0 a b >; s) 0 ab < ; d) 0 a b < ; e) ; 2 0 ab > f) ; g) 2 0 ab < 2 0 a b < ekanligi ma’lum bo’lsa, va bsonlarning ishorasini toping. a 44. ekanligi ma’lum bo’lsa, ifodaning ishorasi qanaqa ? 2 a > a) 3 ; b) 10 ; s) 6 2 a− 5a − 2 a − ; d) (2 )(1 aa ) − − ; e) 2 1 a a − − ; f) ; g) 2 (3)(1aa −− ) 5 2 a− − ; h) (1)(2 (5 ) aa a ) − − + . 45. ekanligi ma’lum bo’lsa, ifodaning ishorasi qanday bo’ladi ? 3 a < 37 a) ; b) 12 ; s) 2 2a − 6 8 4a − a − ; d) (5 )(3 aa ) − − ; e) 4 3 a a − − ; f) ; g) 2 (1)( 2aa −− ) 2 3 a − ; h) 1 ( 2)(3 ) a aa− − − . 46. Agar a) ; b) ; s) 1 1 a < 4 a > 4 a < < ; d) ekanligi ma’lum bo’lsa (1 ifoda qaysi ishoraga ega bo’ladi ? 5 a > )( 4 aa −− ) 47. Agar va bo’lsa, u holda 1 a > 1 b > 1 ab a b+ >+ ekanligini isbotlang. 48. Agar va bo’lsa, u holda ab> 2 b < 2 (2) 2 ba b a + >+ ekanligini isbotlang. 49. Agar bo’lsa, u holda 1 ab >> 22 22ab b a ab a b + +> + + ekanligini isbotlang. 50. Agar bo’lsa, u holda 2 ab << 22 22 24 24 ab b a ab a b + +<++ ekanligini isbotlang. 51. Agar 1 bo’lsa, u holda 2 0 ab <<< 222 2 222 ababaabb ab − −−+ + − > ekanligini isbotlang. 52. Agar bo’lsa, u holda abc≥≥ 222()()( ) ab c bc a ca b 0 − +−−−≥ ekanligini isbotlang. 53. 3 sin ( 0) 6 x xx x >− > tengsizlikni isbotlang. 54. Sonlarni taqqoslang. a) ln 2004 ln 2005 va ln 2005 ln 2006 b) va cos(sin 2006) sin(cos 2006) 55. 0 x> uchun 2 12ln x x +≤ tengsizlik o’rinli bo ’lishini isbotlang. 56. 12,,..., n x x x musbat sonlar bo’lsin. 38 1 1 1 ... ,0 () ... , 0 n n n xx f n xxαα α α α α ⎧⎛⎞ ++ ⎪ ≠ ⎪ ⎜⎟ = ⎨ ⎝⎠ ⎪ ⋅ ⋅= ⎪ ⎩ funtsiyaning monoton o’suvchi bo’lish ini isbotlang. Bundan tashqari f funktsiya qat’iy o’suvchi bo’ladi, faqat va faqat shu holdaki, qachonki jx sonlarning hammasi o’zaro teng bo’lmasa. 57. 33 sin sin sin 8 αβγ ≤ tengsizlikni isbo tlang, bu yerda , α β va γ biror uchburchakning ichki burchaklari. 58. lar 12,,..., n aa a 1 (0 : ) ( 1,..., ) 2 kak n ∈= 12 ... 1 n aa a va + ++ = xossalarga ega bo’lgan sonlar bo’lsin. 2 22 2 12 11 1 1 1 ... 1 ( 1) n n n aa a ⎛⎞ ⎛⎞⎛⎞ −− −≥− ⎜⎟ ⎜⎟⎜⎟ ⎝⎠⎝⎠ ⎝⎠ ekanligini isbotlang. 59. Ixtiyoriy musbat sonlar uchun , , abc 22 22 22 3 3 22 2 ab bc ac a b c abc cabbcac +++ ++≤ + + ≤ + + 3 ab tengsizlik bajarilishini isbotlang. 60. Agar 1 bo’lsa, u holda abc <≤≤ 33 1111 () ln ln ln 3 ln ln ln ln ln ln abc abcabc abc a b c a b c c b c bc ac ab ⎛⎞ ++≤++ ++ ≤++ ++ ⎜⎟ ⎝⎠ 3 ? bo’lishini isbotlang. 61. 23 232,(2,3,4,...nn n −++≥ = ) ekanligini isbotlang. 62. Ixtiyoriy nomanfiy sonlar uchun , , abc ( )( )( ) 8abbcca abc +++≥ tengsizlik o’rinli bo’ lishini isbotlang. 39 63. Ixtiyoriy musbat sonlar uchun 12,,..., n aa a 12 2 3 11 34 1 2 ... 2 nn n aa a a a a a a n aa a a − ++ ++ +++ +≥ tengsizlik o’rinli bo’ lishini isbotlang. Yensen tengsizligi: 64. ()() 22 22 12 3 (, 0) ii iii i aa ab abab a + ++≤+ ∑∑∑ > tengsizlikni isbotlang. (Ko’rsatma: 2 1 yx =+ ). 65. 1 1 n i i i a Sa n = ≥ −− ∑ n tengsizlikni isbotlang, bu yerda 12 ... , 0 ni Sa a a a = ++ > . 66. tengsizlikni isbotlang. 1 11 () , 1, nn pp p ii ii xn xpx − == ≥⋅ > > ∑∑ 0 i 2 36 c Koshi-Bunyakovskiy tengsizligi: 67. tengsizlikni isbotlang, bu yerda lar uchburchakning tomonlari; lar uchburchakning shu tomonlarga tushirilgan balandliklari; uchburchakning yuzi. 222222 ()( ) abc abchhh S++ ++ ≥ , , abc ,,abhhh S 66. 22 (1 ) (1 ) 1, 1, 1 ab a b a b +− −≤ ≤ ≤ ekanligini isbotlang . 68. Agar bo’lsa, 2 3 14 abc ++= 222 14 abc + +≥ bo’lishini isbotlang. Koshi tengsizligi: 69. nomanfiy sonlar uchun , , abc 222 1 abc + += shart bajariladi. 3 abc ++≤ ekanligini isbotlang. 70. b e r i l g a n . ,, 0, 1 abc a b c ≥++= (1 ) , (1 ) (1 ) 8 , 1abcabcabc − −−≥ ++= tengsizlikni isbotlang. 71. Isbotlang: ( )( )( ), (1 )(1 )8 , 1 abc a b c a c b b c a b c abc a b c ≥ +− +− +− − − ++= 40 72. , , 0, 1x y z xyz≥ = berilgan. (3 2 )(3 2 )(3 2 ) 216 x yz y zx z xy + +++++≥ ni isbotlang. Bernulli tengsizligi: 75. 22 44 111 1 xxx x −+++−++= 4 tenglamani yeching. 76. 4411 xx −+ += 4 tenglamani yeching. 77. 46 6 4 1111 32 4 36 x x x ⎛⎞⎛ −+ += − + + ⎜⎟⎜ ⎝⎠⎝ x⎞ ⎟ ⎠ tenglamani yeching. 78. 10 10 10 10 444 4 1111 () () ( ) ( ) ( ) ab c d abcd bcd a abcd ++ + ≥ +++ tengsizlikni isbotlang. 41 Manbaalar ro’yxati 1. Hojoo Lee. Topics in Inequalities-Theorems and Techniques. Seoul: 2004. 2. Andreescu T., Dospinescu G ., Cirtoaje V., Lascu M. Old and new inequalities. Gil Publishing House, 2004. 3. Mathematical Olympiads, Problems a nd solutions from around the world, 1998- 1999. Edited by Andreescu T. and Feng Z. Washington. 2000. 4. Math Links, http://www.mathlinks.ro 5. Art of Problem Solving, http://www.artofproblemsolving.com 6. Math Pro Press, http://www.mathpropress.com 7. K.S.Kedlaya, A<B, http://www.unl.e du/amc/a-activities/a4-for-students/s- index.html 8. T.J.Mildorf, Olympiad Inequalities, http://web.mit.edu/tmildorf 9. Алфутова Н .Б ., Устинов А.В . Алгебра и теория чисел . Сборник задач для математических школ . М .: МЦНМО , 2002. 10. А. Engel. Problem-Solving Strategies. Part s 1,2 . Springer-Verlag New York Inc. 1998. 11. Ayupov Sh., Rihsiyev B., Quchqorov O. «Ma tematika olimpiadalar masalalari» 1,2 qismlar. T.: Fan, 2004 12. Математические задачи , http://www.problems.ru 13. Беккенбах Э ., Беллман Р . Неравенства . — М.: Мир , 1965. 14. Беккенбах Э ., Беллман Р . Введение в неравенства . — М.: Мир , 1965. 15. Коровкин П . П . Неравенства . — Вып. 5. — М.: Наука , 1983. 16. «Математика в школе » (Россия ), « Квант» (Россия ), «Соровский образовательный журнал » (Россия ), “Crux mathematicor um with mathematical Mayhem” ( Канада), “Fizika, matematika va informatika” ( Ўзбекистон) журналлари . 42

8 -MAVZU . SONLI TENGSIZLIKLAR VA ULARNING XOSSALARI. TENGSIZLIKLARNI ISBOTLASH. Reja. 1. Sonli tengsizliklar 2. Sonli tengsizlik larning asosiy xossalari 3. O’rtacha qiymatlar va ular orasidagi munosabatlar 4. Umumlashgan Koshi tengsizligi. 5. Umumlashgan Yung tengsizligi 6. Gel’der tengsizligi. Kalit so ’zlar : Sonli tengsizlik , o’rtacha qiymatla r, Koshi tengsizligi , Yung tengsizligi , Gel’der tengsizligi 1. Sonli tengsizliklar va ularning asosiy xossalari Sonlarni taqqoslash amaliyotda keng qo’llaniladi. Masalan, iqtisodchi rejada ko’zda tutilgan ko’rsatkichlarni amaldagi ko’rsatkichlar bilan taqqos -laydi, shifokor bemorning haroratini sog’lom kishining harorati bilan taqqos laydi, chilangar yo’nayotgan buyumining o’lchamlarini andaza bilan taqqos -laydi. Bu uchala holda qandaydir sonlar o’zaro taqqoslanadi. Sonlarni taqqos -lash natijasida sonli tengsizliklar hosil bo’ladi. 1.1.1 -Ta’rif. Agar &#3627408462; − &#3627408463; ayirma musbat bo’lsa, u holda &#3627408462; son &#3627408463; sondan katta deyiladi. Agar &#3627408462; − &#3627408463; ayirma manfiy bo’lsa, u holda &#3627408462; son &#3627408463; sondan kichik deyiladi. Agar &#3627408462; son &#3627408463; sondan katta bo’lsa, bu &#3627408462; > &#3627408463; kabi; agar &#3627408462; son &#3627408463; sondan kichik bo’lsa, bu &#3627408462; < &#3627408463; kabi yoziladi. Shunday qilib, &#3627408462; > &#3627408463; tengsizlik &#3627408462; − &#3627408463; ayirma musbat, ya’ni &#3627408462; − &#3627408463; > 0 ekanini bildiradi, &#3627408462; < &#3627408463; tengsizlik esa &#3627408462; − &#3627408463; < 0 ekanini bildiradi.

2. Sonli tengsizlik larning asosiy xossalari

1.1 -m iso l. Agar &#3627408475; &#3627408474; to’g’ri kasr bo’lsa, u holda &#3627408475; &#3627408474; < &#3627408475;+1 &#3627408474;+1 bo’lishini isbotlang. ∆ &#3627408475; &#3627408474; kasr &#3627408475; < &#3627408474; bo’lganda (n va m – natural sonlar) to’g’ri kasr deb ata lishini eslatib o’tamiz. Ushbu &#3627408475; &#3627408474; − &#3627408475;+1 &#3627408474;+1= &#3627408475;(&#3627408474;+1)−&#3627408474;(&#3627408475;+1) &#3627408474;(&#3627408474;+1) = &#3627408475;−&#3627408474; &#3627408474;(&#3627408474;+1) ayirma noldan kichik , chunki &#3627408475; − &#3627408474; < 0,&#3627408474; > 0,&#3627408474; + 1 > 0. Binobarin, &#3627408475; &#3627408474; < &#3627408475;+1 &#3627408474;+1 .▲ 1. 2-m iso l. Agar &#3627408462; ≠ &#3627408463; bo’lsa, u holda &#3627408462;2+ &#3627408463;2 > 2&#3627408462;&#3627408463; bo’lishini isbotlang. ∆ &#3627408462;2+ &#3627408463;2− 2&#3627408462;&#3627408463; ayirma musbat ekanligini isbotlaymiz. Haqiqatan ham, &#3627408462;2+ &#3627408463;2− 2&#3627408462;&#3627408463; = (&#3627408462; − &#3627408463;)2 > 0, chunki &#3627408462; ≠ &#3627408463; . ▲ 1. 1-teorema . Agar &#3627408462; > &#3627408463; va &#3627408463; > &#3627408464; bo’lsa, u holda &#3627408462; > &#3627408464; bo’ladi. Δ Shartga ko’ra &#3627408462; > &#3627408463; va &#3627408463; > &#3627408464;. Bu &#3627408462; − &#3627408463; > 0 va &#3627408463; − &#3627408464; > 0 ekanini bildiradi. &#3627408462; − &#3627408463; va &#3627408463; − &#3627408464; musbat sonlarni qo’shib, (&#3627408462; − &#3627408463;)+ (&#3627408463; − &#3627408464;)> 0 ni hosil qilamiz, ya’ni &#3627408462; − &#3627408464; > 0. Demak, &#3627408462; > &#3627408464;. ▲

1.2 -teorema. Agar tengsizlikning ikkala qismiga ayni bir son qo’shilsa, u holda tengsizlik ishorasi o’zgarmaydi. Δ &#3627408514; > &#3627408515; bo’lsin . Bu holda ixtiyoriy &#3627408464; son uchun &#3627408462; + &#3627408464; > &#3627408463; + &#3627408464; teng -sizlikning bajarilishini isbotlash talab qilinadi. Ushbu (&#3627408462; + &#3627408464;)− (&#3627408463; + &#3627408464;)= &#3627408462; + &#3627408464;− &#3627408463; − &#3627408464; = &#3627408462; − &#3627408463;ayirmani qaraymiz. Bu ayirma musbat, chunki masalaning shartiga ko’ra &#3627408462; > &#3627408463;. Demak, &#3627408462; + &#3627408464; > &#3627408463; + &#3627408464;. ▲ 1.1 -Natija. Istalgan qo’shiluvchini tengsizlikning bir qismidan ikkinchi qis -miga shu qo’shiluvchining ishorasini qarama -qarshisiga almashtirgan holda ko’chirish mumkin. Δ &#3627408514; > &#3627408515; + &#3627408516; bo’lsin . Bu tengsizlikning ikkala qismiga – &#3627408464; sonni qo’shib, &#3627408462; − &#3627408464; > &#3627408463; + &#3627408464; − &#3627408464; ni hosil qilamiz, ya’ni &#3627408462; − &#3627408464; > &#3627408463; ▲ 1. 3-teorema. Agar tengsizlikning ikkala qismi ayni bir musbat songa ko’ paytrilsa, u holda tengsizlik ishorasi o’zgarmaydi. Agar tengsizlikning ikkala qismi ayni bir manfiy songa ko’paytrilsa,u holda tengsizlik ishorasi qarama -qarshisiga o’zgaradi. Δ 1) &#3627408514; > &#3627408515; va &#3627408516; > ?????? bo’lsin . &#3627408462;&#3627408464; > &#3627408463;&#3627408464; ekanini isbotlaymiz. Shartga ko’ra &#3627408462; − &#3627408463; > 0 va &#3627408464; > 0. Shuning uchun (&#3627408462; − &#3627408463;)&#3627408464; > 0, ya’ni &#3627408462;&#3627408464; − &#3627408463;&#3627408464; > 0. Demak, &#3627408462;&#3627408464; > &#3627408463;&#3627408464; . 2) &#3627408514; > &#3627408515; va &#3627408516; < ?????? bo’lsin . &#3627408462;&#3627408464; < &#3627408463;&#3627408464; ekanini isbotlaymiz. Shartga ko’ra &#3627408462; − &#3627408463; > 0 va &#3627408464; < 0. Shuning uchun (&#3627408462; − &#3627408463;)&#3627408464; < 0, ya’ni &#3627408462;&#3627408464; − &#3627408463;&#3627408464; < 0. Demak, &#3627408462;&#3627408464; < &#3627408463;&#3627408464; . ▲ 1. 2-Natija. Agar tengsizlikning ikkala qismi ayni bir musbat songa bo’linsa, u holda tengsizlik ishorasi o’zgarmaydi. Agar tengsizlikning ikkala qismi ayni bir manfiy songa bo’linsa,u holda tengsizlik ishorasi qarama -qarshisiga o’zga -radi.

1.4. -teorema. Bir xil ishorali tengsizliklarni qo’shishda xuddi shu ishorali tengsizlik hosil bo’ladi: agar &#3627408462; > &#3627408463; va &#3627408464; > &#3627408465; bo’lsa, u holda &#3627408462; + &#3627408464; > &#3627408463; + &#3627408465; bo’ladi. Δ Shartga ko’ra &#3627408462; − &#3627408463; > 0 va &#3627408464;− &#3627408465; > 0. Ushbu ayirmani qaraymiz: (&#3627408462; + &#3627408464;)− (&#3627408463; + &#3627408465;)= &#3627408462; + &#3627408464; − &#3627408463; − &#3627408465; = (&#3627408462; − &#3627408463;)+ (&#3627408464;− &#3627408465;). Musbat sonlarning yig’indisi musbat bo’lgani uchun (&#3627408462; + &#3627408464;)− (&#3627408463; + &#3627408465;)> 0,ya’ni (&#3627408462; + &#3627408464;)> (&#3627408463; + &#3627408465;). ▲ 1. 5. -teorema. Chap va o’ng qismlari musbat bo’lgan bir xil ishorali teng -sizliklarni ko’paytirish natijasida xuddi shu ishorali tengsizlik hosil bo’ladi: agar &#3627408462; > &#3627408463;,&#3627408464; > &#3627408465; va &#3627408462;,&#3627408463;,&#3627408464;,&#3627408465;– musbat sonlar bo’lsa , u holda &#3627408462;&#3627408464; > &#3627408463;&#3627408465; bo’ladi. Ushbu ayirmani qaraymiz: &#3627408462;&#3627408464; − &#3627408463;&#3627408465; = &#3627408462;&#3627408464; − &#3627408463;&#3627408464; + &#3627408463;&#3627408464; − &#3627408463;&#3627408465; = &#3627408464;(&#3627408462; − &#3627408463;)+ &#3627408463;(&#3627408464;− &#3627408465;) Shartga ko’ra &#3627408462; − &#3627408463; > 0,&#3627408464; − &#3627408465; > 0,&#3627408463; > 0,&#3627408464; > 0. Shuning uchun (&#3627408462; − &#3627408463;)+ &#3627408463;(&#3627408464;− &#3627408465;)> 0, ya’ni &#3627408462;&#3627408464; − &#3627408463;&#3627408465; > 0, bundan &#3627408462;&#3627408464; > &#3627408463;&#3627408465; . ▲

O'xshashlar

Tasodifiy miqdorlar va ularning sonli xarakteristikalari
Tengsizliklarni isbotlashning ba’zi usullari
Logarifmik tenglamalar va tengsizliklar
FUNKSIONAL TENGLAMA VA TENGSIZLIKLARNI O‘RGANISH METODIKASI
IKKI O’ZGARUVCHILI TENGSIZLIKLAR SISTEMASINI YECHISH USULLARI