How to show $\forall n \in \mathbb{N^*}, (\frac{2n}{3}+\frac{1}{3})\sqrt{n} \leq \sum_{k=1}^{n}\sqrt{k}$?

Question

How to show by induction

$\forall n \in \mathbb{N^*}, (\frac{2n}{3}+\frac{1}{3})\sqrt{n} \leq \sum_{k=1}^{n}\sqrt{k}$

Thanks for heping me :)

Answer 1

For $n=1$, the inequality is $1 \leq 1$, so it is true.

Suppose the inequality to be true for an integer $n$, i.e. that $$\left( \frac{2n+1}{3}\right)\sqrt{n} \leq \sum_{k=1}^n \sqrt{k}$$

Then one has $$\sum_{k=1}^{n+1} \sqrt{k} = \sqrt{n+1} + \sum_{k=1}^n \sqrt{k} \geq \sqrt{n+1} + \left( \frac{2n+1}{3}\right)\sqrt{n} $$

Now, it is easy to see that $(2n+1)\sqrt{n} \geq 2n \sqrt{n+1}$. Indeed, squaring it, this inequality is equivalent to $(2n+1)^2n \geq 4n^2(n+1)$, i.e. $4n^3+n+4n^2 \geq 4n^3 + 4n^2$ which is obviously true.

So you get $$\sum_{k=1}^{n+1} \sqrt{k} \geq \sqrt{n+1} + \frac{2n}{3} \sqrt{n+1} = \frac{2(n+1)+1}{3} \sqrt{n+1}$$

You got the inequality at the rank $n+1$.