# Prove by induction $\sqrt{1}+\sqrt{2}+\sqrt{3}+…+\sqrt{n}\ge\frac{2}{3}n\sqrt{n}$ for all positive integers

Assumption: $\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}\ge\frac{2}{3}k\sqrt{k}$

Prove true for $n=k+1$ $$\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}+\sqrt{k+1}\ge\frac{2}{3}(k+1)\sqrt{k+1}$$

I'm upto :$$\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}+\sqrt{k+1}\ge\frac{2}{3}k\sqrt{k}+\sqrt{k+1}$$after which I'm stuck.

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.
• I expanded the right-hand side and then subtracted $2\sqrt{k+1}$ on both sides. An edit has been made showing this. – James Harrison Sep 21 '14 at 0:34