How to prove $a_1^m + a_2^m + \cdots + a_n^m \geq \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}$ I was asked to prove an inequality:
For any $n$ positive numbers $\{a_i\}$ with $a_{1}a_{2}\cdots a_{n} = 1$ and $m \geq n-1$ be a non-negative integer,
$a_1^m + a_2^m + \cdots + a_n^m \geq \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}$
After several attempts, I think I had to use induction on both $m$ and $n$. In particular, I fix an $n$ and claim that the inequality holds for every $m$, and use induction on $n$.
The case where $n = 2$ could be done by rewriting the inequality as $a_1^2 + a_2^2 \geq a_1 + a_2$ and assume  that $a_1 \leq 1 \leq a_2$. 
In the inductive step, given $a_1^k + a_2^k \geq a_1 + a_2$ , by writing $a_1^{k+1} + a_2^{k+1} = (a_1^{k} + a_2^{k})(a_1 + a_2) - a_1^{k}a_2^{k} (a_1 + a_2)$, the result follows.
However, this method seems to fail for $n>2$, as the right-hand side of the inequality becomes more complicated. Should I try not to prove it by induction?
 A: Hint: apply power mean inequality many times. 
Do you see how to do the $m=n-1$ case? Just take $n-1$ terms on the LHS and apply AM-GM. 
A: We will prove that 
$$a_1^m + a_2^m + \cdots + a_n^m \ge a_1^{n-1} + a_2^{n-1} + \cdots + a_n^{n-1} \geq \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}.$$
Left-hand side inequality:
Obviously we have $(x^{n-1}-1)(x^{m-n+1}-1) \ge 0$ for any $x>0$ (note that $m\ge n-1$). Expanding we get
$$x^m\ge x^{n-1} + (x^{m-n+1}-1).$$
Applying this inequality for $x=a_1,\ldots,a_n$ and taking the sum we get
$$\sum_{i=1}^n a_i^m \ge \sum_{i=1}^n a_i^{n-1} + \left(\sum_{i=1}^n a_i^{m-n+1}- n \right).$$
But by AM-GM inequality
$$\sum_{i=1}^n a_i^{m-n+1} \ge n\times (a_1a_2\cdots a_n)^{(m-n+1)/n} = n.$$
Thus we have
$$\sum_{i=1}^n a_i^m \ge \sum_{i=1}^n a_i^{n-1}.$$
Right-hand side inequality:
Denote $S=a_1^{n-1} + a_2^{n-1} + \cdots + a_n^{n-1}$. Using AM-GM inequality for $n-1$ numbers we have
$$S-a_1^{n-1} = a_2^{n-1} + \cdots + a_n^{n-1} \ge (n-1)a_2\cdots a_n = \frac{n-1}{a_1},$$
i.e.
$$S-a_1^{n-1} \ge \frac{n-1}{a_1}.$$
Similarly, we have the other inequalities for $a_2,\ldots,a_n$. Taking the sum of these $n$ inequalities we get
$$nS-(a_1^{n-1} + a_2^{n-1} + \cdots + a_n^{n-1}) \ge (n-1)\left(\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}\right).$$
The left-hand side of the last inequality is just $(n-1)S$. Divided both sides by $(n-1)$ we get the desired inequality.
