<s>[INST] Inequality proof by induction, what to do next in the step I have to prove that for $n = 1, 2...$ it holds: $2\sqrt{n+1} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}}$
Base: For $n = 1$ holds, because $2\sqrt{2}-2 < 1$
Step: assume holds for $n_0$.
$2\sqrt{n+2} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n + 1}}$. But I do not know what to do next? How this can be proved?
 [/INST] You assume it holds for $n$, i.e.
$$ 2\sqrt{n+1} - 2 < 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}}. $$
You now want to prove that it holds for $n+1$, i.e.
$$ 2\sqrt{n+2} - 2 < 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}}. $$
Subtracting the first inequality from the second, you only have to show that
$$ 2(\sqrt{n+2}-\sqrt{n+1}) < \frac{1}{\sqrt{n+1}}. $$
Rearrangement yields
$$ \frac{\sqrt{n+2} - \sqrt{n+1}}{\sqrt{n+1}} = \frac{\sqrt{n+2}}{\sqrt{n+1}} - 1 < \frac{1}{2}, $$
so
$$ \frac{\sqrt{n+2}}{\sqrt{n+1}} < \frac{3}{2}. $$
Squaring both sides gives us
$$ \frac{n+2}{n+1} < \frac{9}{4}, $$
after which cross multiplication results in
$$ 4n+8 < 9n+9, $$
which is obvious for all $n \geq 1$.
</s>