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Tengsizliklar-III
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34.
1
1 , , 2 2 3 a b c z x y = = = deb belgilash kiritsak, u holda 2 2 2 ( , , ) 2 6 12
a b c = + + ifodaning eng katta qiymatini topsak masala yechiladi. , ,
{ }
1 max , ,
2 a b c < ≤ (1) 2 3 2 6
c a ac + ≥ (2) 2 5 2 10 c b bc + ≥ (3) (2) dan
2 2 2 2 2 2 2 3 2 3 1 2 3 2 6
12 2 6 a a a c a c a c ⎛ ⎞ + ≥ ⇒ + ≥ ⇒ + ≥ ⎜ ⎟ ⎝ ⎠ , Bundan
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 3 1 2 3 1 5 2 1 1 6 6 2 6 a a a c a c a a c a a c c a c c ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ + = + − ≤ + + − ≤ + + − = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Xuddi shunday (1) va (3) dan 2 2 7 10
c + ≤ ekanligini topamiz. Shunday qilib, 2 2
2 2 118 ( , , ,) 2( ) 6(
) 4 15
a c b c c = + + + + ≤
Tenglik 1 1 1 118 ( , , )
, , 15 3 5 2 Q a b c Q ⎛ ⎞ = = ⎜ ⎟ ⎝ ⎠ bajariladi va 1 1 1 , , 3 5 2 a b c = = =
qiymatlar (1)-(2)-(3) shartlarni qanoatlantiradi. Bundan 118
max ( , , ) max ( , , ) 15
Q a b c = = .
,
a b x b c y c a t + =
+ = + = belgilash kiritsak yuqoridagi tengsizlik quyidagi ko’rinishga keladi: 2 2 2 2 2 2 2 2 2 2( ) 2( ) 2( ) 8 (*)
( ) 2 ( ) 2 ( ) 2 x t y t x y x t y y y t x x x y t t + + + + + ≤ + −
+ + −
+ + −
+
Ushbu 2 2 2 2( ) (
) t p t p + ≥ + tengsizlikdan foydalansak, 35
( ) ( ) ( ) ( ) ( ) 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 2 2 2 2 2 2 4 4 4( ) (2(
) 2 ) 2
(2( ) 2 ) 2 (2( ) 2 ) 2 4 4 4 ( ) 2 ( ) 2 ( ) 2 4 4 4 4 4 4 2 2 1 1 2 1 2
1 1 1 ( ) ( ) ( ) y t x y x t x t y y y y t x x x x y t t t x t y t x y x t y y t x x y t y x t y x t x t y t x y x t y t x y + + + + + ≤ + −
+ + + − + + + − + + + + + ≤ + + = + + + + + + = + + ≤ + + = + + + + + + + + + + + + ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 4 4 8.
t y t x y x y t x y t x y t + + + = + + = + + + + + +
Bundan (*) isbotlandi. 36. O’rta arifmetik va o’rta geometrik miqdorlar haqidagi Koshi tengsizligini quyidagi usulda qo’llaymiz: 0,6
0,6 0,6
1 1 1 0,36( 0,64)
0,36( 0,64)
0,36( 0,64)
1 1 1 1,2 1,2
1 1 1 1 1 1 a b c b c a a b c b c a + + ≥ + +
+ + + +
⎛ ⎞ ⎜ ⎟ ≥ + + = ⎜ ⎟ ⎜ ⎟ + + + +
+ + ⎝ ⎠ Chunki
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ( 1) ( 1) 1 1 1 1
bc c ac a a b c b c a ac a ac a ac ab b a bc c ac a a ac ac a ac a + + = + + = + +
+ + + +
+ + + +
+ + = + + = + + = + + + + + +
+ + + + + +
36
37. Bu tengsizlikni shakl almashtirish natijasida 1 1
0 ( 1) ( 1) ( 1) y z x z x y x y y z z x ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ + + + − + − + − ≥ ⎜ ⎟ ⎜ ⎟ ⎜
⎟ + + + ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ yoki 3 1 1 1 x z x y y z x y z + + + + + ≥ + + + (*)ga teng kuchli tengsizlikka olib kelamiz. Koshi-Bunyakovskiy-Shvarts tengsizligiga ko’ra ( )( ) 1
x xy x z + ≤
+ +
munosabat o’rinli. Bundan 2 ( 1) (1 ) x x z x y + + ≥ + yoki
1 1 (1 ) x z x x x y + + ≥ + + munosabatni hosil qilamiz. Demak 3 1
1 1 1 1 (1 ) (1 ) (1 ) 1 1 1 3 3 (1 ) (1 ) (1 ) x z x y z y x y z x y z x y y z z x x y z x y y z z x + + + + + + + + ≥ + + ≥ + + + + + + + + + ≥ ⋅ ⋅ = + + +
38. Avval
0 t ∀ > uchun 2 3
3 3( 1) 1 (*) t t t t − +
≥ + + tengsizlik o’rinli ekanligini ko’rsatamiz. (*) ni shakl almashtirib, quyidagi munosabatni hosil qilamiz: 4 2 ( 1) (2
2) 0 t t t − − + ≥ bu tengsizlik 0
∀ > uchun o’rinli. Bundan 2 3 2 3 2 3 6 3 6 3 6 3 3( 1) 3( 1) 3( 1) ( 1)( 1)(
1) (**) x x y y z z x x y y z z − +
− + − +
≥ + + + + + +
munosabatni hosil qilamiz. Umumlashgan Koshi-Bunyakovskiy-Shvarts tengsizligini qo’llasak, 6 3 6 3 6 3 2 2 2
3 ( 1)( 1)( 1) (
1) (***) x x y y z z x y z xyz + + + + + + ≥ + + (**) va (***) larni hadma-had ko’paytirib, isboti talab etilgan tengsizlikni hosil qilamiz.
O’rta arifmetik va o’rta geometrik miqdorlar haqidagi Koshi tengsizligini qo’llaymiz: 37 1 1 1 1 1 2 3
1 1 2 3
1 1 1 1 ...
1 1 ... 1 ...
2 3 2 3 4 2 3 4 1 . n n n n n n n n n n n n n − − − ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ + − + −
+ + − = + + + + + ≥ ⋅ ⋅ ⋅
= ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝
⎠ ⎝ ⎠ = =
40. Koshi-Bunyakovskiy-Shvarts tengsizligini quyidagi usulda qo’llab, 1 1
1 1 1 x y z x y z x y z x y z x y z − − − − +
− + − ≤
+ + ⋅ + + = + +
munosabatni hosil qilamiz.
Berilgan tengsizlikdan 3 3 3 2 2 2
2 2 2 2 ( )( ) 9 ( )
b c a b c a b c a b c + + + + ≥ = + +
tengsizlikni hosil qilamiz. Bu tengsizlik esa Koshi - Bunyakovskiy tengsizligiga ko’ra o’rinlidir.
O’rta arifmetik va o’rta geometrik miqdorlar o’rtasidagi munosabatni quyidagi usulda qo’llaymiz: 2 2
2 2 2 2 2 2 2 2 3 4 2 2 2 2 2 2 2 1 1 1 1 1 9 1 1 1 8 1
1 1 ( ) 9 (
) 9 4 9 8 1 4 3 8 3 9 (
) 3 9 ( ) 3 ( ) 28 3
9 ( ) x y t x y z x y z x y z x y z x y z x y z x y z x y z xyz xyz x y z xyz x y z ⎛ ⎞ ⎛ ⎞ + + + = + + + + + + ≥ ⎜ ⎟ ⎜ ⎟ + + + + ⎝ ⎠ ⎝ ⎠ ≥ + ≥ + = + +
+ + + +
= + +
= −
belgilash kiritsak, u holda ( ) 2 2 2 2 3 2 3 1 1 1 1 2 2 1 1
n n n n k k k k k k k k d n a n n n a a n n n n = = = = = − = ⋅
− + ≤ − ⋅ + + = ∑ ∑ ∑ ∑
munosabatni hosil bo’ladi. Bundan 1 k d ≤ yoki
1 1
n a n − ≤
≤ + .
38
Tengsizlikni ikkala qismini a b c + + ga bo’lib va , ,
b c x y z a b c a b c a b c = = = + +
+ + + +
belgilash kiritsak ( )( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1
1. 2 2 2 2 x y z x y z xy yz zx xy yz zx x y z α β γ α β γ αβ βγ γα
γα αβ βγ α β γ
+ + + + + + + ≤ + + + + + + + + + + + + = + + + + + = + =
45. O’rta arifmetik va o’rta geometrik miqdorlar haqidagi Koshi tengsizligini quyidagi usulda qo’llaymiz: 3 2 2 3 3 3 3 3
3 3 1 1 1 1 1 1 1 1 1 1 2 1 8 ( ) 2 3 3 3 3 3 ( ) 64 3( ) a ab a b b a b ab a b a b ab ab ab a b ⎛ ⎞ + + + = + + + ≥ + = ≥ ⎜ ⎟ ⎝ ⎠ ≥ +
46.
[ ] 1 2 6 , .... 0;1
x x x ∈ ekanligidan tengsizlikning chap qismi quyidagi ifodadan kichik yoki teng 3 3 3 1 2 6 5 5 5 5 5 5 5 5 5 1 2 6 1 2 6 1 2 6 3 3 3 1 2 6 5 5 5 1 2 6 ...
..... 4 ...... 4 .....
4 ...
3 .(*)
... 4 5
x x x x x x x x x x x x x x x x x x + + + = + + + + + + + + + + + + + + + = ≤ + + + +
Ixtiyoriy 0
≥ uchun ( ) ( ) 2 5 3 3 2 3 2 5 1 3 6 4 2 0 t t t t t t + ≥
⇔ − + + + ≥ munosabat o’rinli ekanligidan foydalansak (*) kelib chiqadi.
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