Demonstrate that $(x+y)\ln \left(\frac{x+y}{2}\right) \leq x\ln x +y\ln y$

Question

I need to demonstrate this statement:

$$(x+y)\ln \ \left(\displaystyle\frac{x+y}{2}\right) \leq x\ln x +y\ln y \quad \forall \ x,y \in \ ]0,+\infty[$$

I noticed that $\quad\displaystyle\frac 12x + \frac 12y \quad $is a convex linear combination but I wasn't able to use this property to prove the statement. I also figured out that: $$x< \frac{x+y}{2}<y \quad (\text{or }y< \frac{x+y}{2}<x) \quad \text{and}$$ $$\ln x< \ln \left(\frac{x+y}{2}\right)< \ln y \quad (\text{or }\ln y< \ln \left(\frac{x+y}{2}\right)<\ln x)$$

But again I'm not able to take advantage of these facts.
This problem should be solved with only calculus and real-analysis knowledge. Any hints, tips or solution will be appreciated, thanks in advance.

Answer 1

$f(x)=x\ln{x}$ is a convex function because $$f''(x)=(\ln{x}+1)'=\frac{1}{x}>0.$$

Thus, by Jensen we obtain $$\frac{x\ln{x}+y\ln{y}}{2}\geq\frac{x+y}{2}\ln\frac{x+y}{2}.$$