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?

  • 4
    $\begingroup$ If $a_i$ are positive integers with $a_1a_2\ldots a_n=1$ then you have $a_i=1$ for all $1\leq i\leq n$... $\endgroup$ Commented Jun 13, 2014 at 2:28
  • 1
    $\begingroup$ True, of course, but it seems pretty clear that the typeset letter combination "integers" should be read and pronounced "reals" in that particular sentence :-) $\endgroup$
    – fedja
    Commented Jun 13, 2014 at 3:38
  • $\begingroup$ Of course what I meant is positive numbers instead of positive integers :P $\endgroup$
    – Leon
    Commented Jun 13, 2014 at 9:41
  • $\begingroup$ @Ivan Thanks for correcting the typo ;) $\endgroup$
    – Leon
    Commented Jun 13, 2014 at 9:43

2 Answers 2


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.


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .