= Solution
A function $f$ on an interval is a <convex function> if
$$
f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)
\qquad(0\leq t\leq1).
$$
<Jensen inequality> states that for an integrable random variable $X$, when all terms are defined,
$$
f(\mathbb E X)\leq\mathbb E[f(X)].
$$
The function $\phi(x)=x\log x$ is convex on $(0,\infty)$ because
$$
\phi''(x)=\frac1x>0.
$$
Applying Jensen's inequality to the uniform distribution on the positive numbers $x_1,\ldots,x_n$ gives
$$
\frac1n\sum_{i=1}^nx_i\log x_i
\geq
\left(\frac1n\sum_{i=1}^nx_i\right)
\log\left(\frac1n\sum_{i=1}^nx_i\right).
$$
Writing $S=\sum_i x_i$ and multiplying by $n/S$ yields
$$
\boxed{
\frac{\sum_{i=1}^nx_i\log x_i}{\sum_{i=1}^nx_i}
\geq\log\left(\frac{\sum_{i=1}^nx_i}{n}\right)}.
$$
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