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$\sum_\limits{k=1}^{n-1}\frac{1}{\sqrt{k(n-k)}}\geq 1$ for all $n\geq 2$

Basecase n=2

$\sum_\limits{k=1}^{2-1}\frac{1}{\sqrt{k(2-k)}}=1\geq 1$

Assumption

$\sum_\limits{k=1}^{n-1}\frac{1}{\sqrt{k(n-k)}}\geq 1$ holds for some $n$

Claim

$\sum_\limits{k=1}^{n}\frac{1}{\sqrt{k(n+1-k)}}\geq 1$ holds too

Step Assume $n$ is odd

$\sum_\limits{k=1}^{n}\frac{1}{\sqrt{k(n+1-k)}}=\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{2(n-1)}}+...+\frac{1}{\sqrt{n}}$

Then, to determine which of those terms is the smallest, we want to find the maximum of $k(n+1-k)$.

The deriviative would be $n+1-2k$. So $k=\frac{n+1}{2}$, that's why we assume $n$ is odd in this step.

So our smallest term looks like: $\frac{1}{\sqrt{(\frac{n+1}{2})^2}}=\frac{2}{n+1}$

And since we add this term $n$ times, the sum is bounded below by $\frac{2}{n+1}n=\frac{2n}{n+1}$.

Via induction it is very easy to see, that $\frac{2n}{n+1}>1$ for all n>1, which is all we care about.

Now, how do I proceed for even $n$?

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  • $\begingroup$ Are you interested in a different proof? $\endgroup$ Dec 31, 2016 at 1:23
  • $\begingroup$ @OlivierOloa please only give the approach :) $\endgroup$
    – SAJW
    Dec 31, 2016 at 1:26
  • $\begingroup$ You have now two interesting answers below. $\endgroup$ Dec 31, 2016 at 1:27
  • 2
    $\begingroup$ @OlivierOloa hehe, yes that went quick :) $\endgroup$
    – SAJW
    Dec 31, 2016 at 1:27
  • $\begingroup$ @OlivierOloa Only saw your comment after posting my answer, didn't mean to spoil it. $\endgroup$
    – dxiv
    Dec 31, 2016 at 1:30

2 Answers 2

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Using $\sqrt{ab}\leq \frac{a+b}{2}$ it follows that $\frac{1}{\sqrt{k(n-k)}}\geq \frac{2}{n}$, therefore lhs $\geq \frac{2(n-1)}{n}\geq1$ for $n\geq 2$.

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By AM-GM $\sqrt{k(n-k)} \le \frac{1}{2}\big(k+(n-k)\big) = \frac{n}{2}\,$ Then:

$$\sum_{k=1}^{n-1}\frac{1}{\sqrt{k(n-k)}}\geq \sum_{k=1}^{n-1}\frac{2}{n} = \frac{2(n-1)}{n} \ge 1 \;\;\text{for}\;\; n \ge 2$$

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