Abdullayev jonibekning matematik analiz fanidan yozgan
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koshi tengsizligi va uning tadbiqlari
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... − + + + + + + ≤ + + + + + + + + + ≤ + + + + + + p n p n n p p p n n n p n n p p p n n n m m m x m x m x m x m x m x m m m m x m x m x m m m m x m x m x m
bo’ladi. Bu tengsizlikda , 1 − = p p q
, ...,
, , 2 2 1 1 q n n q q b m b m b m = = =
1 1
2 2 1 1 1 1 ..., , , − − − = = = q n n n q q b a x b a x b a x
desak, q p q n q q p n p p p n n b b b a a a b a b a b a ) ... )( ...
( ) ... ( 2 1 2 1 2 2 1 1 + + + + + + ≤ + + +
q n q q p p n p p n n b b b a a a b a b a b a 1 2 1 1 2 1 2 2 1 1 ) ... ( ) ... ( ... + + + + + + ≤ + + +
Gyo’lder tengsizligi kelib chiqadi.
MASALA YECHISH NAMUNALARI.
1-Misol. Perimetri p 2 bo’lgan uchburchaklar orasida yuzasi eng katta bo’lgan uchburchakni topish so’ralsin. Yechish: Buning uchun Geron formulasi va Koshi tengsizligidan foydalanamiz:
.
3 3 ) ( ) ( ) ( ) )( )( ( 2 3
c p b p a p p c p b p a p p S = − + − + − ⋅ ≤ − − − =
Demak, perimetri p 2 bo’lgan ixtiyoriy uchburchakning yuzasi 3 3 2 p dan oshmaydi. 3 3
p ga faqat ,
− = − = − ya’ni c b a = = bo’lganda teng bo’ladi. Bu esa, bir xil perimetrli uchburchaklar orasida yuzasi eng katta bo’ladigan teng tomonli uchburchak bo’lishini bildiradi. 2-Misol. Ixtiyoriy uchburchak uchun ushbu
4 1 1 1 3 3 3 ≥ + + + + +
tengsizlik o’rinli bo’lishini ko’rsatamiz. Buning uchun berilgan tengsizlik chap tomonini guruhlab yozib olamiz va Koshi tengsizligini qo’llaymiz: p c b a c c b b a a c c b b a a 4 2 2 2 1 2 1 2 1 2 1 1 1 3 3 3 3 3 3 = + + = = ⋅ + ⋅ + ⋅ ≥ + + +
+ +
Bu yerda tenglik faqat
c c b b a a 1 , 1 , 1 3 3 3 = = = bo’lganda, ya’ni 1 = = =
b a bo’lganda bajariladi. 3-Misol. ) 0
, 3 2 2 6 2 > + + + =
x x x x y funksiyaning eng kichik qiymatini topamiz. Buning uchun berilgan funksiyani ushbu
2 2 2 6 2 1 1 1 1 1 x x x x x x x y + + + + + + = ko’rinishida yozib olamiz va Koshi tengsizligini qo’llaymiz: . 7 1 1 1 1 1 7 7 2 2 2 6 2 = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ≥
x x x x x x y
Bu yerda tenglik 1 = x bo’lganda bajariladi. Demak, berilgan funksiyaning eng kichik qiymati 7 ekan. 4-Misol. Agar 0 ..., , 0 , 0 5 2 1 > > > a a a bo’lsa, ushbu 2 5
1 5 1 5 4 5 4 3 4 3 2 3 2 1 ≥ + + + + + + + + + a a a a a a a a a a a a a a a
tengsizlik bajarilishini isbotlaymiz. 2 7 1 6 ,
a a a = = belgilash kiritib, berilgan tengsizlikni ushbu
2 5 7 6 5 6 5 4 5 4 3 4 3 2 3 2 1 ≥ + + + + + + + + + a a a a a a a a a a a a a a a
ko’rinishda yozib olamiz va Yensen tengsizligidan foydalanamiz. Yensen tengsizligini x x f 1 ) ( = funksiya uchun yozamiz:
, ...
... ...
... 2 1 2 2 1 1 1 2 1 2 2 1 1
n n n n n m m m x m x m x m m m m x m x m x m + + + + + + ≤ + + + + + + − . ... ) ...
( ...
2 2 1 1 2 2 1 2 2 1 1
n n n n x m x m x m m m m x m x m x m + + + + + + ≥ + + +
Oxirgi tengsizlikda 5 , 4 , 3 , 2 , 1 , , , 5 2 1 = + = = = + +
a a x a m n i i i i i desak, ushbu
( ) ( ) ( ) ( ) ( ) ( 7 6 5 6 5 4 5 4 3 4 3 2 3 2 1 2 5 4 3 2 1 7 6 5 6 5 4 5 4 3 4 3 2 3 2 1
a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a + + + + + + + + + + + + + ≥ ≥ + + + + + + + + +
tengsizlik kelib chiqadi. Endi esa quyidagi tengsizlikni isbotlashga o’tamiz: , 2 5 ) ( ) ( ) ( ) ( ) ( ) ( 7 6 5 6 5 4 5 4 3 4 3 2 3 2 1 2 5 4 3 2 1 ≥ + + + + + + + + + + + + + a a a a a a a a a a a a a a a a a a a a
)], ( ) ( ) ( ) ( ) ( [ 5 ) ( 2 7 6 5 6 5 4 5 4 3 4 3 2 3 2 1 2 5 4 3 2 1
a a a a a a a a a a a a a a a a a a a + + + + + + + + + ≥ ≥ + + + +
, ) ... ( 2 5 4 5 3 4 3 5 2 4 2 3 2 5 1 4 1 3 1 2 1 2 5 2 2 2 1 a a a a a a a a a a a a a a a a a a a a a a a + + + + + + + + + ≥ ≥ + + + , ) ( ) ( 5 2 5 4 3 2 1 2 5 2 4 2 3 2 2 2 1
a a a a a a a a a + + + + ≥ + + + +
). 1 1 1 1 1 )( ( ) ( 2 2 2 2 2 2 5 2 4 2 3 2 2 2 1 2 5 4 3 2 1 + + + + + + + + ≤ + + + + a a a a a a a a a a
Bu tengsizlik o’rinli, chunki u Koshi-Bunyokovskiy tengsizligidir . Demak 0 ..., , 0 , 0 5 2 1 > > > a a a uchun 2 5
1 5 1 5 4 5 4 3 4 3 2 3 2 1 ≥ + + + + + + + + + a a a a a a a a a a a a a a a tengsizlik o’rinli. Isbot tugadi.
FOYDALANILGAN ADABIYOTLAR:
1. Hasanov.A.B, Yaxshimurotov A.B “Koshi tengsizlikligi va uning tadbiqlari” Urganch-2003 2. Mirzaahmedov M.A, Sotiboldiyev D.A “ O’quvchilarni matematik olimpiadalarga tayyorlash” Toshkent-1993 3. Azlarov T, Mansurov H “Matematik analiz” 1-qism Tosh:1994 4. Nazarov X, Ostonov K “Matematika tarixi” Toshkent-1996 5.Toxirov.A, Mo’minov.F “Matematika olimpiada masalalari” Tosh:1996
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