$$\begin{array}{l} \text{In the lemma, let } a_{11}=a_{3}^{m}, a_{12}=a_{13}=\cdots \\ =a_{\text{ct}}=1, (i=1,2, \cdots, n) \\ \text{It is easy to see: } a_{1}^{m n}+a_{2}^{m}+\cdots+a_{n}^{m} \\ \geqslant \frac{1}{n^{m-1}}\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{m} \\ \text{Then }\left(a_{1}+\frac{k}{a_{1}}\right)^{m}+\left(a_{2}+\frac{k}{a_{2}}\right)^{m}+\cdots+\left(a_{a}+\right. \\ \left.\frac{k}{a_{n}}\right)^{m} \\ \geqslant \frac{1}{n^{n-1}}\left[\left(a_{1}+\frac{k}{a_{1}}\right)+\left(a_{2}+\frac{k}{a_{2}}\right)+\cdots+\left(a_{n}+\right.\right. \\ \left.\left.\frac{k}{a_{n}}\right)\right]^{\mathrm{mm}} \\ =\frac{1}{n^{m-1}}\left[\left(a_{1}+a_{2}+\cdots+a_{n}\right)+k\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots\right.\right. \\ \left.\left.+\frac{1}{a_{n}}\right)\right]^{m} \\ \geqslant \frac{1}{n^{\frac{m}{n}-1}}\left[k+k \cdot \frac{n^{2}}{a_{1}+a_{2}+\cdots+a_{n}}\right]^{m} \\ =\frac{1}{n^{k-1}}\left(k+n^{2}\right)^{\mathrm{kx}} \text{. } \\ =n\left(n+\frac{k}{n}\right)^{n x} . \end{array}$$