I want to show that for all $x_i > 0$: $$\sum_{i=1}^{n}\dfrac{x_i}{x_1 + \ldots+x_n}\;\log(x_i) \geq \log\left(\dfrac{x_1 + \ldots + x_n}{n}\right)$$
I thought of Jensen's inequality but since $\log$ is a concave function, this seems to give me the opposite as for $x_i > 0, a_i \geq 0$, it holds: $a_1\log(x_1) + \ldots + a_n\log(x_n) \leq \log(a_1x_1 + \ldots + a_nx_n)$.
Or is there some algebraic manipulation we can do on the original inequality such that we can use Jensen (or any other inequality for that matter)?
The inequality can be rewritten to $$ \frac{x_1 + \ldots + x_n}{n} \cdot \log\left(\frac{x_1 + \ldots + x_n}{n}\right) \le \frac 1n \sum_{i=1}^n x_i \log(x_i) \, , $$ and that is true because the function $f(x) = x \log(x)$ is convex.