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