13. Given $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, and $a_{1}+a_{2}+\cdots+a_{n}=1, k$ is a positive integer, find the minimum value of $\frac{1}{a_{1}^{k}\left(1+a_{1}\right)}+\frac{1}{a_{2}^{k}\left(1+a_{2}\right)}+\cdots+\frac{1}{a_{n}^{k}\left(1+a_{n}\right)}$.
13. From $a_{1}+a_{2}+\cdots+a_{n}=1$, we get $1=a_{F}+a_{2}+\cdots+a_{n} \geqslant n \sqrt[n]{a_{1} a_{2} \cdots a_{n}}$, thus $\frac{1}{\sqrt[n]{a_{1} a_{2} \cdots a_{n}}} \geqslant n^{n}$. For $k$ being a positive integer, taking $k$ instances of $\left(a_{1}+a_{2}+\cdots+a_{n}\right)$, by the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(a_{1}+a_{2}+\cdots+a_{n}\right) \cdots\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left[\left(1+a_{1}\right)+\left(1+a_{2}\right)+\cdots+\right. \\
\left.\left(1+a_{n}\right)\right]\left[\frac{1}{a_{1}^{k}\left(1+a_{1}\right)}+\frac{1}{a_{2}^{k}\left(1+a_{2}\right)}+\cdots+\frac{1}{a_{n}^{k}\left(1+a_{n}\right)}\right] \geqslant \\
\left(\sqrt[k+2]{a_{1}^{k}} \cdot \sqrt[k+2]{1+a_{1}} \cdots \cdots \sqrt[k+2]{\frac{1}{a_{1}^{k}\left(1+a_{1}\right)}}+\sqrt[k+2]{a_{2}^{k}} \cdot \sqrt[k+2]{1+a_{2}} \cdots \cdots\right. \\
\sqrt[k+2]{\frac{1}{a_{2}^{k}\left(1+a_{2}\right)}}+\cdots+\sqrt[k+2]{a_{n}^{k}} \cdot \sqrt[k+2]{1+a_{n}} \cdots \cdots \\
\left.\sqrt[k+2]{\frac{1}{a_{n}^{k}\left(1+a_{n}\right)}}\right)^{k+2}=n^{k+2}
\end{array}$$

That is,
$$(n+1) y \geqslant n^{k+2}$$

Therefore,
$$y \geqslant \frac{n^{k+2}}{n+1}$$
The minimum value of $\frac{1}{a_{1}^{k}\left(1+a_{1}\right)}+\frac{1}{a_{2}^{k}\left(1+a_{2}\right)}+\cdots+\frac{1}{a_{n}^{k}\left(1+a_{n}\right)}$ is $\frac{n^{k+2}}{n+1}$. (For $k=1, n=3$, this is the problem from the 6th issue of 2004 in "Middle School Mathematics")