China Western Mathematical Olympiad

Let $y_i = \frac{1}{1+x_i}$, $i=1, 2, \dots, 5$, then $x_i = \frac{1-y_i}{y_i}$, $i=1, 2, \dots, 5$ and $\sum_{i=1}^5 y_i = 1$.
We have
$$\begin{align*} \sum_{i=1}^{5} \frac{x_i}{4+x_i^2} &\le 1 \Leftrightarrow \sum_{i=1}^{5} \frac{-y_i^2+y_i}{5y_i^2-2y_i+1} \le 1 \\ &\Leftrightarrow \sum_{i=1}^{5} \frac{-5y_i^2+5y_i}{5y_i^2-2y_i+1} \le 5 \\ &\Leftrightarrow \sum_{i=1}^{5} \left(-1+\frac{3y_i+1}{5y_i^2-2y_i+1}\right) \le 5 \\ &\Leftrightarrow \sum_{i=1}^{5} \frac{3y_i+1}{5\left(y_i-\frac{1}{5}\right)^2+\frac{4}{5}} \le 10. \end{align*}$$
Furthermore,
$$\begin{align*} \sum_{i=1}^{5} \frac{3y_i + 1}{5\left(y_i - \frac{1}{5}\right)^2 + \frac{4}{5}} &\le \sum_{i=1}^{5} \frac{3y_i + 1}{\frac{4}{5}} = \frac{5}{4} \sum_{i=1}^{5} (3y_i + 1) \\ &= \frac{5}{4} \times (3+5) = 10. \end{align*}$$
This completes the proof.