Inequality involving a finite sum

The following statements are equivalent. The first statement shows that $P(n)<2\sqrt{n}\Rightarrow P(n+1)<2\sqrt{n+1}$ and the last statement is evidently true:

$2\sqrt{n}+\frac{1}{\sqrt{n+1}}\leq2\sqrt{n+1}$

$2\sqrt{n\left(n+1\right)}+1\leq2\left(n+1\right)$

$2\sqrt{n\left(n+1\right)}\leq2n+1$

$4n\left(n+1\right)\leq4n^{2}+4n+1$

Add $\frac{1}{\sqrt{k+1}} $ to both sides of $P(k)$ and then show

$$2\sqrt{k}+\frac{1}{\sqrt{k+1}} < 2\sqrt{k+1}$$

First note that $$\frac{1}{\sqrt{n}} = \frac{2}{\sqrt{n}+\sqrt{n} }\leq \frac{2}{\sqrt{n}+\sqrt{n-1}} \overset{(1)}{=} 2\left(\sqrt{n}-\sqrt{n-1} \right)$$ $(1)$ follows by multiplication with conjugate.

Now sum to get: $$\sum_ {k=1}^{n} \frac{1}{\sqrt{k}} \leq \sqrt{n}-\sqrt{0}= \sqrt{n}$$