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.

  • $\begingroup$ I don't quite get how you got your second line $\endgroup$ – user117498 Sep 21 '14 at 0:31
  • 1
    $\begingroup$ I expanded the right-hand side and then subtracted $2\sqrt{k+1}$ on both sides. An edit has been made showing this. $\endgroup$ – James Harrison Sep 21 '14 at 0:34

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