<s>[INST] Prove the following inequality by induction for all $n\in \Bbb N$ I know how to prove the base case for this. But how would I continue from there?

 [/INST] Set $$S_n =\sum_{i=1}^n \frac{1}{\sqrt{i}} \geq \sqrt{n}$$
then  $$S_{n+1} =\sum_{i=1}^{n+1} \frac{1}{\sqrt{i}} \geq \sqrt{n} =S_n +\frac{1}{\sqrt{n+1}}$$
Obviously $S_1\ge 1 =\sqrt1.$
If we assume that $$S_n\ge \sqrt{n}$$
then
we have that 
$$S_{n+1} =S_n +\frac{1}{\sqrt{n+1}} \ge \sqrt{n}+\frac{1}{\sqrt{n+1}} =\frac{\sqrt{n^2+n} +1}{\sqrt{n+1}} $$
But we know that, 
$$n^2\le 4n^2 +4n \Longleftrightarrow n\le 2 \sqrt{n^2+n}\Longleftrightarrow n+\color{blue}{n^2+n+1} \le 2 \sqrt{n^2+n}+\color{blue}{n^2+n+1} $$
that $$(n+1)^2 = n^2 +2n +1 \le 2 \sqrt{n^2+n}+\color{blue}{n^2+n+1} = (\sqrt{n^2+n}+1)^2$$
we just prove that, $$\sqrt{n^2+n}+1\ge n+1 \Longleftrightarrow \frac{\sqrt{n^2+n} +1}{\sqrt{n+1}}\ge \sqrt{n+1}$$
Hence, 
$$S_{n+1}\ge \sqrt{n}+\frac{1}{\sqrt{n+1}} =\frac{\sqrt{n^2+n} +1}{\sqrt{n+1}} \ge \sqrt{n+1}$$
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