I need help proving an inequality by induction 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.
 A: 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.
A: 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
$$2(k+1)\sqrt{k+1}<(k+1)\sqrt{k+1}+(k+1)\sqrt{k+2}$$
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.
