How do I prove by induction that $\displaystyle\sum_{n=1}^k\sqrt{n}>\frac23k\sqrt{k}$ ?

I believe it has something to do with the property $2\sqrt{k}\le\sqrt{k}+\sqrt{k+1}\le2\sqrt{k+1}$,

but I could not crack it.


We need to show that \begin{eqnarray*} \frac{2}{3} k \sqrt{k} +\sqrt{k+1} > \frac{2}{3} (k+1) \sqrt{k+1}. \end{eqnarray*} Rearranging gives \begin{eqnarray*} 2 k \sqrt{k} &>& (2k-1) \sqrt{k+1} \\ 4 k^3 &>& (4k^2-4k+1)(k+1) \\ 3k &>& 1 \end{eqnarray*} Which is obvious.

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  • $\begingroup$ I have seen this "proof" before, and I do not like it. I would like a proof from beginning to end (not the other way around) using the property I mentioned above. $\endgroup$ – יהודה שמחה ולדמן Nov 25 '19 at 12:04
  • $\begingroup$ You are essentially trying to show the inequality by using $\sqrt{k} < \sqrt{k+1}$. I cannot see an easy way to use this to reach $2 k \sqrt{k} > (2k-1) \sqrt{k+1}$. $\endgroup$ – Donald Splutterwit Nov 25 '19 at 12:12
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    $\begingroup$ @יהודהשמחהולדמן you are right that this isn't a proof until you turn it around. However, turning it around is not hard to do. But please, please don't be that guy who demands "You have to prove it the exact way I failed to prove it!". $\endgroup$ – Paul Sinclair Nov 25 '19 at 20:20

For the induction step we are assuming $\sum_{n=1}^k\sqrt{n}>\frac23k\sqrt{k}$ and we need to prove that this implies $\sum_{n=1}^{k+1}\sqrt{n}>\frac23(k+1)\sqrt{k+1}$ and we can proceed as follows

$$\sum_{n=1}^{k+1}\sqrt{n}=\sqrt{k+1}+\sum_{n=1}^{k}\sqrt{n}\,\stackrel{Ind. Hyp.}>\,\sqrt{k+1}+\frac23k\sqrt{k} \,\stackrel{?}> \,\frac23(k+1)\sqrt{k+1}$$

and conclude how already indicated by Donald Splutterwit's answer.

As an alternative, following your idea, we can try to use that


we obtain

$$\frac23(k+1)\sqrt{k+1} <\frac13 (k+1)\sqrt{k+1}+\frac13(k+1)\sqrt{k+2}$$

and need to prove that

$$\frac13 (k+1)\sqrt{k+1}+\frac13(k+1)\sqrt{k+2}<\sqrt{k+1}+\frac23k\sqrt{k}$$

$$ (k+1)\sqrt{k+1}+(k+1)\sqrt{k+2}<3\sqrt{k+1}+2k\sqrt{k}$$

which is not true.

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