Prove that $x\log x$ is midpoint convex.

Question

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.

Answer 1

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