A consequence of the convexity of $ f(x) = x \log x $ I verified that $f:\mathbb{R_{+}^{*}} \rightarrow \mathbb{R}, f(x) = x \log x $ is convex, since it is twice differentiable and $f''(x) = \frac{1}{x}$ is positive for the domain. But my teacher asked to verify the following consequence:
$ \forall a,b,x,y > 0 ,$    $x \log \frac{x}{a} + y \log \frac{y}{b} \geq (x+y) \log \frac{x+y}{a+b} $
But I'm failing miserably at seeing why convexity would imply such inequality.
$f$ being convex isn't $f((1-t)x+ty) \leq (1-t)f(x) + tf(y), \forall t \in [0,1]$?
So I would have
$((1-t)x+ty) \log ((1-t)x+ty)  \leq (1-t)x \log x + ty \log y, \forall t \in [0,1]$
But I can't see how $a$ and $b$ are related to $t$.
Can someone help me?
Thanks.

Consider $x/a=X$ and $y/b=Y$. Then
$$
\frac{x+y}{a+b}=\frac{aX+bY}{a+b}=\frac{a}{a+b}X+\frac{b}{a+b}Y
$$
so we can set $t=b/(a+b)$ and $1-t=a/(a+b)$. Then
$$
\frac{x+y}{a+b}=(1-t)X+tY
$$
and, by convexity,
$$
\frac{x+y}{a+b}\log\frac{x+y}{a+b}\le
(1-t)f(X)+tf(Y)=
\frac{a}{a+b}\frac{x}{a}\log\frac{x}{a}+\frac{b}{a+b}\frac{y}{b}\log\frac{y}{b}
$$
that is,
$$
(x+y)\log\frac{x+y}{a+b}\le x\log\frac{x}{a}+y\log\frac{y}{b}
$$
