Prove the following logarithm inequality. 
If $x, y \in (0, 1)$ and $x+y=1$, prove that $x\log(x)+y\log(y) \geq \frac {\log(x)+\log(y)} {2}$.  

I transformed the LHS to $\log(x^xy^y)$ and the RHS to $\log(\sqrt{xy})$, from where we get that $x^xy^y \ge \sqrt{xy}$ beacuse the logarithm is a monotonically increasing function. From there we can transfrom the inequality into $x^{x-y}y^{y-x} \ge 1$. So here I am stuck.  
I could have started from some known inequalities too, like the inequalities between means.

This can also be viewed as a special case of Chebyshev's sum inequality.
For any increasing function $f:I \to \Bbb R$ and $x, y \in I$ we have
$$
 \frac 12 \bigl( x f(x) + y f(y) \bigr) \ge \left( \frac 12(x+y) \right) \cdot \left( \frac 12(f(x)+f(y)) \right) \\
\implies x f(x) + y f(y) \ge \frac{x+y}{2} (f(x) + f(y)) \, .
$$
More generally, if $x_1, \ldots, x_n \in I$ then
$$
\sum_{k=1}^n x_k f(x_k) \ge \frac{1}{n} \sum_{k=1}^n x_k \sum_{k=1}^n f(x_k) \, .
$$
