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Tengsizliklar-III
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47. Tengsizlikni chap qismini S bilan belgilab, Koshi-Bunyakovskiy-Shvarts tengsizligini quyidagi usulda qo’llab, ( )
) 2 2 2 2 2 2 2 1 2
(1 2 ) (1 2 )
( )
bc b ac c ab a b c ⋅ + + + + + ≥ + + yoki
1 1 2
( )
abc a b c ≥ + + +
39 munosabatni hosil qilamiz. 2 2 2 3 ( ) a b c abc a b c + + ≥ + + tengsizlikka ko’ra ( )
2 2 2 1 1 3 2 2 5 1 3 ( ) 1 3 3 S abc a b c a b c ≥ ≥ = + + + + + + ekanligini hosil qilamiz.
Koshi-Bunyakovskiy-Shvarts tengsizligini quyidagi usulda qo’llab, ( )
) ( ) 2 2 2 2 2 2 3 3 3 2 3 1 1 3 3 3 2 1 2 3 1 1 2 1 ( 1) ... 1 2 3 ... ( 1) 2 2 3 ( 1) 2 ...
, 2 2001
... 1
i i i i i i x x x x x i i x x x x x ёки i x x x i i − − − ⎛ ⎞ − ⎛ ⎞ ⋅ ≥ + + + + + + + + −
≥ ⎜ ⎟ ⎜ ⎟ − ⎝ ⎠ ⎝ ⎠ ≥ + + + + ≥ ≤ ≤ + + +
−
ekanligini topamiz. Bundan 2001 2 1 2 1 1 1 1 1 2 ...
2 1 1,999.
... 1 2 2 3
2000 2001 2001
i i i x x x x = − ⎛ ⎞ ⎛ ⎞ ≥ + + + = − > ⎜ ⎟ ⎜ ⎟ + + + ⋅ ⋅ ⋅ ⎝ ⎠ ⎝ ⎠ ∑
Tengsizlikni 1 1 1 3
b a b c b c b c b c a c a b a + + + ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ − + + − + + − + ≥
⎜ ⎟ ⎜
⎟ ⎜ ⎟ + + + ⎝ ⎠ ⎝ ⎠ ⎝
⎠ yoki
2 2 2 3 ( ) ( ) ( ) b ac c ab a bc b c b c a c a b a + + + + + ≥ + + + kurinishda yozamiz.Koshi-Bunyakovskiy- Shvarts tengsizligini quyidagi usulda qo’llasak, ( ) ( ) 2 2 2 3 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 3. ( ) ( ) ( ) b c a c a b b ac c ab a bc b a c b c b c a c a a b b a c c a b a b c b c a b a c c a b b a c c a b a b c + + + + + + + + ≥ + + ≥ + + + + + + + + + ≥ ⋅ ⋅ = + + +
50. Avval tengsizlikni induktsiya metodi yordamida 2 ,
= ∈ sonlar uchun isbotlaymiz. 2 n = da
( )( ) 2 1 2
1 2 1 2 1 2
1 2 1 2 1 1 1 2 0 1 1 1 (
1)( 1)(
1) y y y y y y y y y y y y − − + − = ≥ + + + + + +
2 k n = da tengsizlik o’rinli bo’lsin deb faraz qilsak, u holda 1 2
n + = da 40 1 1 1 1 1 2 2 2 1 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 1 1 1 1 1 1 ...
.... 1 1 1 1 1 1 2 2 2 1 ... 1 ...
1 ....
k k k k k k k k k k k k k k k y y y y y y y y y y y y y y y + + + + + + + + + ⎛ ⎞ ⎛ ⎞ + + + + + + + ≥ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ + + + + + + ⎝ ⎠ ⎝ ⎠ ≥ + ≥ + + + ⋅ munosabat o’rinli va tengsizlik 2 , ,
n k N = ∈ uchun isbotlandi. Endi tengsizlikni n N ∈ uchun isbotlaymiz. Buning uchun 2 ,
k m n k N = > ∈ uchun tengsizlik o’rinli deb faraz etsak va 1 2 1 2 ....
.. n n n m n y y y y y y + + = = = = deb olsak, u holda 1 2
1 2 1 1 1 ...
1 1 1 1 ... ...
1 n n n n n m n m y y y y y y y y y − + + + + ≥ + + + + +
bo’ladi. Bundan yuqoridagi tengsizlik isbotlanadi.
2
i i x α = , 4 2 i π π α ≤
2 2
1 2 2 1 1 cos cos (1 cos
) (1 cos
) n n n i i i i n n i i i i α α α α = = = = ⎛ ⎞ ⎜ ⎟ Π ⎜ ⎟ ≤ ⎜ ⎟ Π −
− ⎜ ⎟ ⎝ ⎠ ∑ ∑
ko’rinishga, yoki 2 2 2 2 2 2 1 2 1 2 1 1 1 ...
1 1 1 1 ...
n n n n tg tg tg tg tg tg α α α α α α + + + ≥ + + + +
ko’rinishga keladi. Agar 2
i tg y α = deb belgilash kiritsak, 1 i y ≥ bo’ladi va 1 2
1 1 1 ... 1 1 1 1 ... n n n n y y y y y y + + ≥ + + + + ni isbotlash kerak bo’ladi.Bu tengsizlik esa 50- masalada isbotlangan.
Tengsizlikni musbat 1 2 , ,..., n x x x sonlar uchun isbotlash yetarli. Koshi- Bunyakovskiy-Shvarts tengsizligini quyidagi usulda qo’llab, 41 1 2 2 2 2 2 2 2 1 1 2 1 2 2 2 2 1 2 2 2 2 2 2 1 1 2 1 .... 1 1 1 ... ...
1 1 1 ... n n n n x x x x x x x x x x x x n x x x x x + + + < + + + + + + + ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟
+ + + ⎜
⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ + + + + + + ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
munosabatni hosil qilamiz. Bundan 2 2 2 1 2 2 2 2 2 2 2 2 2 2 1 1 2 1 2 .. 1 (1 ) (1 ) (1 ... )
n x x x x x x x x x + + + < + + + + + + +
ekanligini ko’rsatsak, yuqoridagi tengsizlik isbotlanadi. ( ) 2 2 2 2 2 2 2 2 2 2 1 2 1 1 1 2 2 2 2 2 1 1 1 (1 ... )(1 ...
) 1 ... 1 1 1 ... 1 ... k k k k k k k x x x x x x x x x x x x x x − − ≤ = + + + +
+ + +
+ + + + = − + + + + + + bo’lgani uchun 2 2
2 2 1 1 1 1 1 1 1 ... 1 ... n k k k n x x x x x = ⎛ ⎞ < − < ⎜ ⎟ + + +
+ + +
⎝ ⎠ ∑ .
3
3 3 9( ) ( )
b c a b c + + ≥ + +
tengsizlikdan foydalanib, ( )
( ) 3 3 3 3 2 2 2 3 3 3 3 2 3 3 3 1 1 1 1 1 1 6 6 6 9 6 6 6 4 3 ( )
( ) 3 1 1 1 1 3 3 3 3 3 3 ( ) 4 3 3 3 3 3 b c a b c a a b c a b c ab bc ca ab bc ac abc c a b abc ab bc ca abc abc ⎛ ⎞ + + + + + ≤ + + +
+ + = ⎜ ⎟ ⎝ ⎠ − + + − − − = + + + = ≤ + + − ≤ =
munosabatni hosil qilamiz.Endi 3 3 1 abc abc ≤ yoki 2 2 2 1 27 a b c ≤ ekanligi ko’rsatamiz: 3 2 1 ( ) ( )( )( ) 3 27 ab bc ca abc ab bc ca + + ⎛ ⎞ = ≤ = ⎜ ⎟ ⎝ ⎠ .
42
1
) bc a b c − = + ekanligidan 2 2 3 2 2 3 3 3 2 3 3 3 3 3 ( ) (1 ) 2 ( ) 2 1 1 1 1 3 3 3 3 2 3 9 abcx a b c c b x a b c c b x ax bc bc bc bc ax bc bc bc ax bc ax bc = ⋅ ⋅ ≤ + + = + − = ⎛ ⎞ + − ⎜ ⎟ ⎛ ⎞ = + − + ≤ + + = + + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎜ ⎟ ⎝ ⎠ munosabatni hosil qilamiz. Xuddi shunday 3 3
1 1 1 3 , 3
3 9 3 9 abcy by ac abcz cz ab ≤ + + ≤ + + tengsizliklar o’rinli.bu tengsizliklarni hadma-had qo’shib isboti talab etilgan tengsizlikni hosil qilamiz.
Tengsizlikni ikkala qismiga umumiy maxraj tanlab, 1 1 1 1 1 1 1 1 1 1 1 n n n n i i i i i i i i a a n a a = = = = − ≥ ⋅ + + ∑ ∑ ∑ ∑ yoki
1 1 1 1 1 1 ( 1) 1 n n n i i i i i i i n a a a a = = = ≥ + + ∑ ∑ ∑
munosabatni hosil qilamiz. Umumiylikni chegaralamasdan 1 2 ... n a a a ≥ ≥ ≥ deb olsak, oxirgi tengsizlik Chebishev tengsizligia ko’ra o’rinlidir.
1 2
.... n a a a k − + + + = deb olamiz. U holda isboti talab etilgan tengsizlikka teng kuchli bo’lgan 0 0 2 2 2 0 0 0 0 1 1 ( ) (
1)( )( ) ( ) ( 1). n n n n n n a k a k a k a k n n n n n k a a k n a k a k k k a a a a n + +
+ + ⋅ ≥ ⋅ ⇔ + − + + ≥ − + + ⇔ + + ≥ −
tengsizlikni hosil qilamiz. 0 0 2 n n a a a a + ≥ ekanligidan ( ) 0 1
k n a a ≥ − tengsizlikni isbotlasak yuqoridagi tengsizlik isbotlandi. 2 1
( 1,2,...., 1)
− + ⋅ ≤ = − tengsizlikka ko’ra 0 1 2 1 1 2 3 ... n n a a a a a a a a − ≤ ≤ ≤ ≤
yoki 0 1 1 2 2 ...... n n n a a a a a a − − ≤ ≤ ≤ munosabatni hosil qilamiz. Bundan, 43 1 2 1 1 1 2 2 1 1 2 2 0 0 0 0 ... ( ) (
) ... 2 1 2 ... 2 2 ... 2 ( 1) 2 n n n n n n n n n K a a a a a a a a a n a a a a a a a a n a a − − − − − = + + +
= + + + + ≥
+ − + + ≥ + + = ⋅ = −
ekanligi kelib chiqadi.
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