This should be completely elementary, if not downright trivial compared to other problems I've seen on this site. I think I'm just being silly. The problem is from Kaczor and Nowak's problem book in analysis. Somehow I made it through undergraduate without ever seeing half the inequalities in this text, and I really need to know them, so I'm trying to get some practice.

Suppose that $a_k$ are positive numbers, where $k$ ranges from $1$ to $n$, and satisfies $\sum a_k \leq 1$. Show that $\sum \frac{1}{a_k} \geq n^2$.

So, a moment of staring at the problem, and the problem screams AM-GM-HM inequality. We can rewrite what we are asked to prove as $$\frac{1}{\sum \frac{1}{a_k}} \leq \frac{1}{n^2}$$

so that the left side is now precisely the harmonic mean of the $a_k$. By AM-GM-HM, this is less or equal to both the harmonic and geometric means. The obvious choice here being to use the arithmetic mean since it's the only thing for which we can say anything about. This gives:

$$\frac{1}{\sum \frac{1}{a_k}} \leq \frac{\sum a_k}{n} \leq \frac{1}{n} $$

Which is weaker than what I was asked to prove - I've only been able to show the original sum is $\geq n$, rather $n^2$.

Can the approach be fixed? Perhaps there is a better way to go?

  • $\begingroup$ Harmonic mean $\dfrac{1}{\sum \frac{1}{a_k}}$ should be $\dfrac{n}{\sum \frac{1}{a_k}}$ $\endgroup$
    – DHMO
    Apr 18, 2017 at 3:53
  • $\begingroup$ @DHMO Of course, if I use the inequality properly, the result is clear. Only mildly embarrassing. I should mail my degree back... (just kidding of course). Thank you though. $\endgroup$ Apr 18, 2017 at 3:54

3 Answers 3


$$\begin{array}{rcl} \displaystyle \sum_{k=1}^n a_k &\le& 1 \\ \displaystyle \dfrac1n \sum_{k=1}^n a_k &\le& \dfrac1n \\ \displaystyle \dfrac n{\sum_{k=1}^n \frac1{a_k}} &\le& \dfrac1n \\ \displaystyle \dfrac{\sum_{k=1}^n \frac1{a_k}}n &\ge& n \\ \displaystyle \sum_{k=1}^n \frac1{a_k} &\ge& n^2 \\ \end{array}$$

With equality when $\forall k \in 1,2,3\cdots, n:a_k = \dfrac1n$.


The HM has an $n$ on top (it's the reciprocal of the AM of the reciprocals). (Think about what happens if all the $a_k$ are equal.)

With this adjustment, your approach works perfectly.


Two additional proofs.

  1. Cauchy-Schwarz.

$$n^2 = (\sum_{k=1}^n \sqrt{a_k} \frac{1}{\sqrt{a_k}})^2\le \sum_{k=1}^n a_k \times \sum_{k=1}^n \frac{1}{a_k}\le \sum_{k=1}^n \frac{1}{a_k}.$$

  1. Calculus.

For any $\lambda>0$ we have

\begin{align*} 2n \le \sum_{k=1}^n (\lambda a_k + \frac{1}{\lambda a_k})&\le \lambda+ \sum_{k=1}^n \frac{1}{\lambda a_k}.\end{align*}


$$\lambda( 2n -\lambda) \le \sum_{k=1}^n \frac{1}{a_k}.$$

Maximum of LHS at $\lambda = n$, and result follows.


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