Tri-Variable Inequality!
Let x , y and z be positive real numbers such that x y z = 1 . What is the minimum value of:
( 1 + y ) ( 1 + z ) x 3 + ( 1 + z ) ( 1 + x ) y 3 + ( 1 + x ) ( 1 + y ) z 3
The answer is 0.75.
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1 solution
Did all the hard work,..... At last it was the Equality case ( x , y , z ) = ( 1 , 1 , 1 ) \cry ¨
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Hahaha! This is the harsh truth of inequalities that the equality case is mostly equal to 1 .
LOL XD.
This is not the complete solution but an outline.
Suppose x ≤ y ≤ z . Then, ( 1 + y ) ( 1 + z ) ≥ ( 1 + z ) ( 1 + x ) ≥ ( 1 + x ) ( 1 + y )
Then, apply Chebyshev's Sum inequality to c y c l i c ∑ ( ( 1 + y ) ( 1 + z ) x 3 . Solving it, we further have to prove:
4 ( x 3 + y 3 + z 3 ) ( 3 + x + y + z ) ≥ 9 ( 1 + x ) ( 1 + y ) ( 1 + z )
Then use AM-GM on the RHS of the above inequality from which we will need to prove 1 2 ( x 3 + y 3 + z 3 ) ≥ ( 3 + x + y + z ) 2
The rest is for you to solve. Happy Solving!!!