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Tengsizliklar-III
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5.
2 2 2 1 2 ...
n t a a a = + + + deb belgilab,
( )
) { } ( ) 1 2 2 3 1 1 1 2 1 3
1 4 1 2 3 2 4 2 3 4 3 1 2 2 2 2 1 2 1 2 ...
... ...
... .....
1 1 ... ... 1 2 2 n n n n n n n n n n 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 t − − + + +
+ ≤ ≤ + + + + + + + + + + + + + + + + + = = + + − + + = −
munosabatlarni hosil qilamiz. Bu yerdan t<1 kelib chiqadi. Koshi-Bunyakovskiy-Shvarts tengsizligini qo’llab, ( )
)( ) 2 2 2 2 2 2 2 1 1 2 2
1 2 1 2 2 ... ... ...
n n n a b a b a b a a a b b b t + + + ≤ + + + + + +
=
yoki 1 1 2 2
... n n a b a b a b t + + + ≤ tengsizlikni topamiz. Bundan 22 ( ) ( ) ( ) (
) ( ) ( ) ( ) 1 1 2 2 2 3 1 1 1 2 2 2 1 2 2 3 1 ... ... 1 1 .... 1 1 1 1 2 2 n n n n n a b a a b a a b a a b a b a b a a a a a a t t t + + + + +
+ = + + + + + + + + ≤ + − = − − + <
6. Berilgan tengsizlikni chap tomonini T bilan belgilab, o’rta arifmetik va o’rta geometrik miqdorlar haqidagi Koshi tengsizligidan foydalanamiz, ya’ni: 1 1 1 (1 )(1 ) (1 )(1 ) (1 )(1 ) 1 1 1 2 1
1 2 1
1 2 1
1 1 3 2 2
bc ac T ab a b bc b c ac a c ab bc ac a b b c a c a b b c a c b a c b c a a b b c a c a c b c b a c a b a b c = + + = + − − + − − + − −
= + + ≤ − − − − − − ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ≤ + + + + + ≤ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ − − − − − − ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ≤ + + + + + = ⎜ ⎟ + + + + + + ⎝ ⎠
Tengsizlikni ikkala qismidagi qavslarni ochib ixchamlasak 2 2 2 2 2 2 3
b c a c b a b c b c a c b a b c a + + + + + ≥ + + + (*) ifoda hosil bo’ladi.Endi , ,
b c x y z b c a = = = deb belgilash kiritsak, u holda (*) tengsizlik 2 2
1 1 1 3 x y z x y z x y z + + + + + ≥ + + + (**) ko’rinishga keladi. 1
= ekanligidan quyidagi ( )
2 2 2 3 x y z x y z x y z + +
+ + ≥ ≥ + + (1) va
1 1 1 3 x y z + + ≥ (2) tengsizliklar o’rinli. (1) va (2) larni hadma-had qo’shib (**) ni hosil qilamiz. Bundan (*) isbotlandi.
1
+ = ekanligidan foydalanib yuqoridagi tengsizlikni quyidagi 23 shaklda yozamiz: ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 3 a b a b a a b a b b a b ≤ + + + + + + + Bundan
2 2 3 3 a b ab a b + ≤ + yoki ( ) ( ) ( ) ( ) 2 3 3 2 2 0 a b a b ab a b a b + − + = − + ≥
9.
2 2 3 3 4 4 ( ) yz y z y z yz y z + = + ≤ + tengsizlik o’rinli, chunki ( ) ( ) 4 3 3 4 3 3 4 4 2 2 2 2 ( ) 0 ( )
y z yz z y z y z x y z xyz y z y z − − + = − − ≥ ⇒ + ≥ + ≥ + yoki
5 5 4 5 2 2 5 4 4 4 4 4 ( ) x x x x y z x x y z x y z ≥ = + + + + + + . Xuddi shunday, 5 4
4 5 2 2 4 4 4 5 2 2 4 4 4 ,
y z z y x z x y z z x y x y z ≥ ≥ + + + + + + + + . Bu tengsizliklarni hadma-had ko’shib, isboti talab qilingan tengsizlikni hosil qilamiz.
Yuqoridagi tengsizlikni quyidagicha yozib olamiz: 2 2
2 2 2 2 2 2 5 2 2 5 2 2 5 2 2 3 x y z y x z x y z x y z x z x z x y + + + + + + + + ≤ + + + + + +
va Koshi-Bunyakovskiy-Shvarts tengsizligini qo’llab, 5 1 5 2 2 2 2 2 2 2 2 2 2 2 2 2 ( )( ) ( ( ) ) ( ) x y z yz y z x yz y z x y z + + + + ≥ + + ≥ + +
yoki 2 2 2 2 2 5 2 2 2 2 2 . x y z yz y z x y z x y z + + + + ≤ + + + + Xuddi shunday, 2 2
2 2 2 2 2 2 2 5 2 2 2 2 2 5 2 2 2 2 2 ,
y z xz x z x y z xy x y x y z x y z z x y x y z + + + + + + + + ≤ ≤ + + + + + + + +
munosabatlarni hosil qilamiz. Bu tengsizliklarni hadma-had qo’shsak, 2 2 2 2 2 2 2 2 2 5 2 2 5 2 2 5 2 2 2 2 2 2 3
y z x y z x y z xy yz zx x y z y x z z x y x y z + + + + + + + + + + ≤ + ≤ + + + + + + + + 24
11 . , ,
α β γ uchburchak burchaklari uchun 2 2
1 cos cos
cos 2cos cos cos α β
α β γ = + + + tenglikdan foydalanib, cos cos
cos cos
cos cos
2 cos cos
cos cos cos cos
cos cos cos cos
cos cos α β γ α β γ β γ γ α α β β γ γ α α β ⋅ ⋅ = + + + ifodani hosil qilamiz . cos cos
cos , , cos cos cos cos
cos cos x y z α β γ β γ γ α α β = = =
deb belgilash kiritib, 3 cos
cos cos
2 α β γ + + ≤ tengsizlikdan foydalansak, 1 1 1 3 2( ) 3 2 4( 2( )) 9
8( ) 9(
2) 4( ) 5(
) 18 x y z xyz xy yz zx x y z xy yz zx xyz xy yz zx x y z x y z x y z + ≤ ⇔ + + ≤ ⇔ + + + + + + ≤ ⇔ + + ≤ + + +
− + +
= + + +
O’rta arifmetik va o’rta geometrik miqdorlar haqidagi Koshi tengsizligidan foydalanib, ( )
) ( ) ( ) ( ) ( ) 2 2 3 2 3 2 2 3 3 2 2 1 1 1 , 1 1 1 , 2 2 2 1 1 1 2 a b a a a a b b b b c c c c c + + + = + + − ≤ + = + − + ≤ + + = + − + ≤
munosabatlarni topamiz. Endi quyidagi tengsizlikni isbotlasak yetarli: ( )( ) ( )( ) ( )( ) ( ) ( ) (
) ( ) ( )( )( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
4 4 4 4 3 2 2 2 2 2 2 3 2 2 2 2 2 2 2 8 72
a b c a b b c c a a c b a c b a b c a b b c c a a b c a b c + + ≥ ⇔ + + + + + + + + + + + ≥ + + + ⇔ + + + + + ≥ + = Bu tengsizlik quyidagi tengsizliklarni hadma-had qo’shishdan hosil qilinadi: ( )
) 4 2 2 2 2 2 2
2 2 2 2 2 2 3 3 3 48, 2
6 24
b c c a abc a b c a b c + + ≥ = + + ≥ = Bulardan isboti talab qilingan tengsizlikni hosil qilamiz. 25
13. Birinchi navbatda 2 2
x y x y + ≥ + tengsizlikni isbotlaymiz. Faraz qilaylik, 2 2 3 x y x y +
+ bo’lsin. 3 4 2 3 x y x y + ≤ + tengsizlikdan foydalansak, farazimizga zid bo’lgan ( ) (
) ( ) 2 3 3 2 4 2 3 2 2 2 x y x x y y x y + ≥ + + + ≥ + tengsizlik hosil bo’ladi. Shuning uchun ( ) 2 2 3 3 4 2 2 3 3 4 3 3 2 3 3 2 2 1 2 1 2.
x y x y x y x y x y x y x y x y + ≥ + ≥ + ⇒ + ≥ + + + ≥ ≥ + + − +
− ⇒ + ≤
Umumiylikni chegaralamasdan a b c ≥ ≥ deb olib, uchburchak tengsiz-ligini qo’llasak, 1 1 2 2
a b a = + + >
⇒ ≤ < va bundan ( ) (*) 2 2 2 2 2 1 2 1 1 1
n n n n n n n n b a b a < + ⇒ = +
+ .
Endi qo’yidagini qaraymiz: 1 .... 2 2 2 n n n n n n n c n c b b c b b c − ⎛ ⎞ + = + + + > + ⎜ ⎟ ⎝ ⎠ (chunki 1 2 n n n cb c − > ). Xuddi shunday, 2
n n c a a c ⎛ ⎞ + > + ⎜ ⎟ ⎝ ⎠ .
Demak, ( ) ( ) 1 1 1 2 2 n n n n n n c c b c a b b a + + + < + + + = . (**) (*) va (**) larni hadma-had qo’shib, isboti talab etilgan tengsizlikni hosil qilamiz.
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