<s>[INST] Prove by induction that $\sqrt{1}+\sqrt{2}+...+\sqrt{n} \geq \frac{2}{3}n\sqrt{n}$ The base step is pretty obvious: $1 \geq \frac{2}{3}$.
Then we assume that $P(k)$ is true for some $k \in \mathbb{Z}^{+}$ and try to prove $P(k+1)$. So I have
$ \sqrt{1}+\sqrt{2}+...+\sqrt{k} + \sqrt{k+1} \geq \frac{2}{3}k\sqrt{k}+\sqrt{k+1}$ 
by the induction hypothesis. But I'm not too sure how to proceed to prove that this is also greater than $\frac{2}{3}(k+1)\sqrt{k+1}$.
Would appreciate any help!
 [/INST] For the induction step, you want to show that:
$$
\frac{2k\sqrt{k} + 3\sqrt{k+1}}{3} \geq \frac{2(k+1)\sqrt{k+1}}{3} \\
2k\sqrt{k} + 3\sqrt{k+1} \geq 2k\sqrt{k+1} + 2\sqrt{k+1}\\
$$
Working backwards:
$$
2k\sqrt{k} + \sqrt{k+1} \geq 2k\sqrt{k+1} \\
4k^2 \times k \geq (4k^2 - 4k+1)(k+1) = 4k^3 - 4k^2 + k + 4k^2 - 4k+ 1 = 4k^3 - 3k + 1
$$
The rest should follow since $k \geq \frac1{3}$, since the induction is over the positive integers.
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