Problem 1431

First we prove Lemma 1 by mathematical induction.

Lemma 1: When $x, y \ge 0$, $n \ge 2$, where $n$ is a positive integer,
$$(x^n + y^n)^2 \le (x^2 + y^2)^n. \quad (1)$$
Proof: When $n = 2$, it clearly holds.
Suppose that when $n = k$ ($k \ge 2$), the conclusion holds, i.e.
$$(x^k + y^k)^2 \le (x^2 + y^2)^k.$$
When $n = k + 1$,
$$\begin{aligned} \text{the right side of (1)} &= (x^2 + y^2)^{k+1} \\ &= (x^2 + y^2)^k (x^2 + y^2) \\ &\ge (x^k + y^k)^2 (x^2 + y^2) \\ &= (x^{2k} + y^{2k} + 2x^k y^k)(x^2 + y^2) \\ &= x^{2k+2} + y^{2k+2} + x^2 y^{2k} + x^{2k} y^2 + 2x^{k+2} y^k + 2x^k y^{k+2} \\ &\ge x^{2k+2} + y^{2k+2} + 2x^{k+2} y^k + 2x^k y^{k+2} \\ &\ge x^{2k+2} y^{2k+2} + 2x^{k+1} y^{k+1} \\ &= (x^{k+1} + y^{k+1})^2. \end{aligned}$$

In (1), let $x = \sqrt[n]{b}$, $y = \sqrt[n]{c}$, and simplify to obtain
$$(\sqrt[n]{b+c})^2 \le (\sqrt[n]{b})^2 + (\sqrt[n]{c})^2. \qquad (2)$$
Then, by mathematical induction, extend (2) to
$$(\sqrt[n]{x_1 + x_2 + \cdots + x_m})^2 \le (\sqrt[n]{x_1})^2 + (\sqrt[n]{x_2})^2 + \cdots + (\sqrt[n]{x_m})^2,$$
where $n \ge 2$, $m$ is a positive integer.

$$\begin{aligned} & \text{Then } \sqrt[n]{\frac{x_1}{S-x_1}} = \sqrt[n]{\frac{x_1}{\sqrt[n]{S-x_1}}} = \frac{2(\sqrt[n]{x_1})^2}{2\sqrt[n]{x_1}\sqrt[n]{S-x_1}} \\ & \ge \frac{2(\sqrt[n]{x_1})^2}{(\sqrt[n]{x_1})^2 + (\sqrt[n]{x_2} + x_3 + \cdots + x_m)^2} \\ & \ge \frac{2(\sqrt[n]{x_1})^2}{(\sqrt[n]{x_1})^2 + (\sqrt[n]{x_2})^2 + \cdots + (\sqrt[n]{x_m})^2}. \end{aligned}$$

Similarly,
$$\sqrt[n]{\frac{x_m}{S-x_m}} \ge \frac{2(\sqrt[n]{x_1})^2}{(\sqrt[n]{x_1})^2 + (\sqrt[n]{x_2})^2 + \cdots + (\sqrt[n]{x_m})^2},$$
$$\sqrt[n]{\frac{x_m}{S-x_m}} \ge \frac{2(\sqrt[n]{x_m})^2}{(\sqrt[n]{x_1})^2 + (\sqrt[n]{x_2})^2 + \cdots + (\sqrt[n]{x_m})^2}.$$

Adding all the above expressions together gives the original inequality.
From the proof process, we know that the equality above holds if and only if two of the $x_i$ are equal and nonzero, while all the others are $0$.