Prove the follwing inequality : (x^2y^2+z^4)3<=(x^3+y^3+z^3)^4

Mauricio Mathis 2022-07-17 Answered
Let x , y , z be three positive numbers. Can you prove the follwing inequality : ( x 2 y 2 + z 4 ) 3 ( x 3 + y 3 + z 3 ) 4
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Answers (1)

eri1ti0m
Answered 2022-07-18 Author has 11 answers
We need to show ( x 2 y 2 + z 4 ) 3 ( x 3 + y 3 + z 3 ) 4 .
By AM-GM, we have ( x 3 + y 3 + z 3 ) 4 ( 2 ( x y ) 3 2 + z 3 ) 4
Let a = x y > 0. Then it is sufficient to show that ( 2 a 3 + z 3 ) 4 ( a 4 + z 4 ) 3
Let t = a z > 0, then we need to show
f ( t ) = ( 2 t 3 + 1 ) 4 ( t 4 + 1 ) 3 0 for t > 0
or f ( t ) = t 3 ( 15 t 9 + 32 t 6 3 t 5 + 24 t 3 3 t + 8 ) > 0
Now, note that for t 1
3 t 6 3 t 5 = 3 t 5 ( t 1 ) and 3 t 3 3 t = 3 t ( t 2 1 )
so f ( t ) = t 3 [ 15 t 9 + 32 t 6 + 3 t 3 ( 1 t 2 ) + 21 t 3 + 3 ( 1 t ) + 5 ] > 0.
Thus f ( t ) > 0 when t>0, and hence the inequality holds.
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