Proving for all integer $n \ge 2$, $\sqrt n < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt n}$ [duplicate]

$$\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt k}+\frac{1}{\sqrt k+1} > \sqrt{k} + \frac{1}{\sqrt {k+1}} = \frac{\sqrt{k(k+1)} + 1}{\sqrt{k+1}} > \frac{\sqrt{k^2} + 1}{\sqrt{k+1}} = \sqrt{k+1}$$

Hint: Note that $$\sqrt{k+1}-\sqrt{k}=\frac{1}{\sqrt{k+1}+\sqrt{k}}<\frac{1}{\sqrt{k+1}}$$

For the last induction step

$$\sqrt{k^2 + k}+1> k +1 \implies \frac{\sqrt{k^2 + k}}{\sqrt{k+1}}+ \frac1{\sqrt{k+1}}> \sqrt{k+1}\implies \sqrt{k} +\frac1{\sqrt{k+1}}> \sqrt{k+1}.$$

Hence

$$\sum_{j=1}^{k} \frac1{\sqrt{j}} > \sqrt{k} \implies \sum_{j=1}^{k+1} \frac1{\sqrt{j}} > \sqrt{k} + \frac1{\sqrt{k+1}} > \sqrt{k+1}.$$

For a simpler proof

$$\sum_{k=1}^n \frac1{\sqrt{k}}> \frac{n}{\sqrt{n}}= \sqrt{n}.$$

For the inductive step, it suffices to show $\sqrt{n+1} - \sqrt{n} < \frac {1}{\sqrt{n+1}} = \frac{\sqrt{n+1}}{n+1}$ since that implies the inequality gets stronger as $n$ increases. This is clear since

$$\sqrt{n+1} - \sqrt{n} = \frac{\sqrt{n+1}}{n+1} + \frac{(-\sqrt{n}\sqrt{n+1})(\sqrt{n+1} - \sqrt{n})}{n+1}$$

The right addend on the right equation is clearly negative.