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.

Answer

Since $x-y$ and $\log x-\log y$ have the same sign, we have
$$
(x-y)(\log x-\log y)\ge 0

$$ or equivalently
$$
x\log x+y\log y\ge y\log x+x\log y.
$$ Hence it holds that
$$
2x\log x+2y\log y\ge (x+y)\log x+(x+y)\log y=\log x+\log y.
$$ This proves
$$
y\log y+x\log x\ge \frac{\log x+\log y}{2}.
$$

Note: As @Martin R pointed out, the result can be generalized to $$
x+y=1\Longrightarrow\; xf(x)+yf(y)\ge \frac{f(x)+f(y)}{2}
$$ for any increasing function $f:(0,1)\to\Bbb R$.