<s>[INST] Proving $x\ln(\frac{x}{a})+y\ln(\frac{y}{b})\geq (x+y)\ln(\frac{x+y}{a+b})$ 
Let a,b,x,y be positive reals. Prove $x\ln(\frac{x}{a})+y\ln(\frac{y}{b})\geq (x+y)\ln(\frac{x+y}{a+b})$

I don't have any olympic background, so I may be missing some standard trick.
The inequality looks closely related to the concavity of $\ln$.
EDIT: the following is wrong as stated by Macavity.
Indeed using points $\frac{x}{a}$,$\frac{y}{b}$, and $\frac{x}{x+y}$,$\frac{y}{x+y}$ as weights, one has $$\frac{x}{x+y}\ln(\frac{x}{a})+\frac{y}{x+y}\ln(\frac{y}{b})\geq \ln(\frac{x^2}{x+y}\frac{1}{a}+\frac{y^2}{x+y}\frac{1}{b})$$
But I can't prove that $\displaystyle\frac{x^2}{x+y}\frac{1}{a}+\frac{y^2}{x+y}\frac{1}{b}\geq \frac{x+y}{a+b}$
 [/INST] use Cauchy-Schwarz inequality we have
$$\left(\dfrac{x^2}{a}+\dfrac{y^2}{b}\right)(a+b)\ge (x+y)^2$$
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