Prove inequality $x\cdot\ln(x) + y\cdot\ln(y) \geq (x+y)\cdot \ln(x+y)$

Question

I have inequality $$x\cdot\ln(x) + y\cdot\ln(y) \geq (x+y)\cdot \ln(x+y)$$ I transformated it to $$2^{x+y}\cdot x^x\cdot y^y\geq(x+y)^{x+y}$$ And I got stuck. Please help

Answer 1

It's obviously wrong! Try $x=y=1$.

By the way, since $f(x)=x\ln{x}$ is a convex function, we obtain: $$\frac{x\ln{x}+y\ln{y}}{2}\geq\frac{x+y}{2}\ln\frac{x+y}{2},$$ which gives $$x\ln{x}+y\ln{y}\geq(x+y)\ln\frac{x+y}{2}$$

Answer 2

Contradictory example: Let $x=e=y$, the inequality claims $$ e+e > (2e)\cdot(1+\ln(2)) $$ which is wrong.

Answer 3

Ok, it's false ...

But the map $f:(0,+\infty)\to\mathbb{R},t\mapsto x\ln(x)$ is convex and hence :

$$\forall(x,y)\in(0,+\infty)^2,\,f\left(\frac{x+y}{2}\right)\le\frac 12\left(f(x)+f(y)\right)$$

That is :

$$\forall(x,y)\in(0,+\infty)^2,\,x\ln(x)+y\ln(y)\ge(x+y)\ln\left(\frac{x+y}2\right)$$