Let $a_1,...,a_n$ be a sequence of positive numbers. Show that

$$(a_1+a_2+\cdots+a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)\geq n^2$$

Hint: Use the fact that for $x>0$ we have $x+(1/x)\geq 2$.

My idea is to use induction. But I can't seem to make it work. On the other hand, I don't see how the hint is going to take me where I need to go.

  • $\begingroup$ Are you familiar with the Arithmetic Mean Geometric Mean Inequality (AM/GM)? $\endgroup$ Aug 28, 2014 at 17:50
  • $\begingroup$ If you mean the inequality $\sqrt{ab}\leq\frac{a+b}{2}$ $\endgroup$ Aug 28, 2014 at 17:52
  • $\begingroup$ I mean the generalization $\frac{x_1+\cdots+x_n}{n}\ge (x_1x_2\cdots x_n)^{1/n}$. $\endgroup$ Aug 28, 2014 at 17:55
  • $\begingroup$ I am NOT familiar with the general form. $\endgroup$ Aug 28, 2014 at 17:56
  • $\begingroup$ No one has mentioned the Cauchy-Schwarz inequality so far. A proof using it is extremely direct, no thinking required at all. $\endgroup$
    – user26486
    Aug 29, 2014 at 2:59

4 Answers 4


Induction is not necessary. Note that:

$$(a_1+a_2+\cdots+a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right) =\\ \frac{a_1}{a_1} + \frac{a_1}{a_2} + \cdots + \frac{a_1}{a_n} + \frac{a_2}{a_1} + \cdots + \frac{a_n}{a_n} = \\ \sum_{i=1}^n\sum_{j=1}^n\frac{a_i}{a_j} = \frac{1}{2} \sum_{i=1}^n\sum_{j=1}^n\left(\frac{a_i}{a_j} + \frac{a_j}{a_i}\right) \stackrel{*}{\geq} \frac{1}{2}\sum_{i=1}^n\sum_{j=1}^n 2 = \frac{1}{2}\sum_{i=1}^n 2n = \frac{1}{2}·2n^2 = n^2$$

Where in $*$ we used the hint $x + \displaystyle\frac{1}{x} \geq 2$ taking $x = \displaystyle\frac{a_i}{a_j}$

  • 2
    $\begingroup$ change $\frac{a_i}{a_j} + \frac{a_j}{a_i}$ to $\left(\frac{a_i}{a_j} + \frac{a_j}{a_i}\right)$ will look better. $\endgroup$
    – mike
    Aug 29, 2014 at 1:46

This is equivalent to:

$$\frac{a_1+a_2+\ldots+a_n}{n}\geq \frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}}$$

Where the LHS is the Arithmetic Mean and the RHS is the Harmonic Mean.

Those are power means defined as:

$$M_p = \left(\frac{1}{n}\sum_{k=1}^na_k^p\right)^{1/p}$$

It is known that $M_p \geq M_q$ iff $p \geq q$.

Note that the Arithmetic Mean is $M_1$ and the Harmonic Mean is $M_{-1}$

  • $\begingroup$ You left to say "You were fast" and upvote :) $\endgroup$
    – pointer
    Aug 28, 2014 at 17:58
  • $\begingroup$ Do you know of a method that utilizes the given hint? $\endgroup$ Aug 28, 2014 at 18:00
  • $\begingroup$ @user165130 check my other answer ;) $\endgroup$
    – Darth Geek
    Aug 28, 2014 at 18:15
  • $\begingroup$ It looks good! Thanks for the help. $\endgroup$ Aug 28, 2014 at 18:21

Induction works just fine with the hint:


  • $\begingroup$ Ah, I think I see it now. Thank you. $\endgroup$ Aug 28, 2014 at 18:14

Clearly is valid for $n=1$. Assume that it holds for $n$ positive numbers. $$\Big((a_1+\cdots+a_n)+a_{n+1}\Big)\Big((\frac{1}{a_1}+\cdots+\frac{1}{a_n})+\frac{1}{a_{n+1}}\Big)=(a_1+\cdots+a_n)(\frac{1}{a_1}+\cdots+\frac{1}{a_n})$$ $$+(\frac{a_1}{a_{n+1}}+\cdots\frac{a_n}{a_{n+1}})+(\frac{a_{n+1}}{a_{1}}+\cdots\frac{a_{n+1}}{a_{n}})+1$$ $$\geq n^2+\Big((\frac{a_{n+1}}{a_{1}}+\frac{a_1}{a_{n+1}})+\cdots(\frac{a_{n+1}}{a_{n}}+\frac{a_n}{a_{n+1}})\Big)+1\geq n^2+2n+1=(n+1)^2.$$ Notice that $x+\frac{1}{x}\geq 2$ as long as $x>0.$


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