Prove that: (is true for any positive x,y):? $x^{x} \cdot y^{y} \geq {(\frac{x + y}{2})}^{x + y}$ $x^x*y^y>=((x+y)/2)^(x+y)$

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$x^{x} \cdot y^{y} \geq {(\frac{x + y}{2})}^{x + y}$ $x^x*y^y>=((x+y)/2)^(x+y)$

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Answer 1 · 2018-02-27T14:51:05

See below.

Consider $f (x) = x ln x$ $f(x) = x ln x$

This function has a convex hypograph because

$f''(x) = 1/x > 0$

so in this case

$f((x+y)/2) le 1/2(f(x)+f(y))$ or

$((x+y)/2)ln((x+y)/2) le 1/2( x ln x + y ln y)$ or

$((x+y)/2)^((x+y)/2) le (x^x y^y)^(1/2)$

and finally squaring both sides

$((x+y)/2)^(x+y) le x^x y^y$