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Prove that negative entropy function $f(x)=x\log x$ is midpoint convex on $(0,+\infty)$.

Attempt. Let $0<x<y$, so: $$f\left(\frac{x+y}{2}\right)\leq \frac{f(x)+f(y)}{2}$$ is equivalent to: $$\left(\frac{x+y}{2}\right)^{\frac{x+y}{2}}\leq x^{\frac{x}{2}} y^{\frac{y}{2}},$$ being equivalent to: $$\left(\frac{x+y}{2}\right)^{x+y}\leq x^x y^y.$$ The last inequality is equivalent to: $$(x+y)^{x+y}\leq (2x)^x (2y)^y.$$ It would be enough to prove $(x+y)^{x}\leq (2x)^x,~(x+y)^{y}\leq (2y)^y$, but the first inequality doesn't seem to hold, though.

Thanks for the help.

Nikolaos Skout
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    Are you allowed to use the fact that $f$ is convex? In such case, midpoint convexity is just a special case of convexity (and is in fact equivalent to convexity for a continuous function on an interval). As for the convexity of $f$, it easily follows from $f''(x) = 1/x > 0$. – Sangchul Lee Feb 13 '19 at 01:42

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Applying Jensen's inequality to $f$ is an easy way. Otherwise, you can apply GM-HM inequality to obtain $$ x^{\frac{x}{x+y}}y^{\frac{y}{x+y}}\ge \frac{1}{\frac{x}{x+y}\frac1{x}+\frac{y}{x+y}\frac1{y}}=\frac{x+y}2. $$

Myunghyun Song
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