Suppose $ a_{1},...,a_{n} > 0 $ , that $m\geq n-1$ is a positive integer, and
$$\prod_{i=1}^{n}a_{i}=1$$ I'd like to show that
$$\sum_{i=1}^{n}a_{i}^m \geq \sum_{i=1}^{n}\frac{1}{a_{i}}$$
(For reference this is a problem from Excursions in Classical Analysis by Hongwei Chen). To me it seems this problem is most amenable to induction by using AM-GM on the inductive step. The $n=1,2$ cases are easy but I'm not entirely sure how to proceed. Heuristically the inequality seems plausible since for large m, those $a_{i}$ that are larger than 1 will dominate, but I'm more or less stuck.
I don't think induction works. You can use Muirhead's Inequality for this problem. In my proof, $m$ is any real number such that $m\geq n-1$.
Note that the sequence $$\textbf{a}:=(m,\underbrace{0,0,\ldots,0}_{n-1\text{ zeros}})$$ dominates (majorizes) the sequence $$\textbf{b}:=\Big(\underbrace{\frac{m+1}{n},\frac{m+1}{n},\ldots,\frac{m+1}{n}}_{n-1\text{ terms}},\frac{m+1}{n}-1\Big)\,.$$ The $\textbf{a}$-mean $M_\textbf{a}(\textbf{x})$ of $\textbf{x}:=(x_1,x_2,\ldots,x_n)$ is $$M_\textbf{a}(\textbf{x})=\frac1n\,\sum_{i=1}^n\,x_i^m\,,$$ whereas the $\textbf{b}$-mean $M_\textbf{b}(\textbf{x})$ of $\textbf{x}$ is $$M_\textbf{b}(\textbf{x})=\frac1n\,\sum_{i=1}^n\,\frac{1}{x_i}\,\prod_{j=1}^n\,x_j^{\frac{m+1}{n}}=\frac{1}{n}\,\sum_{i=1}^n\,\frac{1}{x_i}\,\left(\prod_{j=1}^n\,x_j\right)^{\frac{m+1}{n}}=\frac{1}{n}\,\sum_{i=1}^n\,\frac{1}{x_i}\,.$$ By Muirhead's Inequality, $M_\textbf{a}(\textbf{x})\geq M_\textbf{b}(\textbf{x})$; therefore, $$\frac1n\,\sum_{i=1}^n\,x_i^m\geq \frac{1}{n}\,\sum_{i=1}^n\,\frac{1}{x_i}\,,$$ which is equivalent to the desired inequality. The equality holds if and only if $x_1=x_2=\ldots=x_n=1$.
Remark. This can be proven using only the Weighted AM-GM Inequality. That is, prove that, for $i=1,2,\ldots,n$, we have $$\frac{1}{x_i}\,\prod_{j=1}^n\,x_j^{\frac{m+1}{n}}\leq \sum_{j=1}^n\,\left(\frac{m+1}{mn}\right)\,x_j^m-\frac{1}{m}\,x_i^m\,.$$