Problem 1568

49. Using the AM-GM inequality, we get
$$x_{1}^{n-1}+x_{2}^{n-1}+\cdots+x_{k}^{n-1} \geqslant k \cdot \sqrt[k]{\left(x_{1} x_{2} \cdots x_{k}\right)^{n-1}}$$

From the given conditions and the AM-GM inequality, we get
$$x_{1} x_{2} \cdots x_{2}=x_{1}+x_{2}+\cdots+x_{4} \geqslant k \cdot \sqrt[k]{x_{1} x_{2} \cdots x_{2}}$$

Thus, $x_{1} t_{2} \cdots x_{k} \geqslant h \cdot \frac{x_{1}}{x_{1} x_{2} \cdots x_{h}}$. Therefore,
$$x_{1} x_{2} \cdots x_{k} \geqslant k^{k}$$

Substituting (3) into (1), we get
$$x_{1}^{n-1}+x_{2}^{n-1}+\cdots+x_{1}^{n-1} \geqslant k \cdot \quad k^{n-1}$$

Next, we prove $\sqrt[k-1]{k^{n-1}} \geqslant n$. That is, we need to prove
$$\sqrt[n-1]{n^{k-1}} \leqslant k$$

Given $1 \leqslant k \leqslant n$, noting the root index $n-1$ and $n$ raised to the power of $k-1$, we add $n-n$ ones inside the root and apply the AM-GM inequality to get
$$\sqrt[n]{n^{n-1}}=\sqrt[n-1]{n^{1} \cdot 1} \leqslant \frac{(k-1) n+(n-k) \cdot 1}{n-1}=k$$

Inequality (3) is thus proven.
(1), (2) hold with equality if and only if $x_{1}=x_{2}=\cdots=x_{i}$. (5) holds with equality if and only if $n=k$. Therefore, the original inequality holds with equality if and only if $n=k \mathrm{H}_{1} x_{1}=x_{2}=\cdots=x_{2}$.