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?

Let's prove this first: x>0, : $ 2*\sqrt{x} +\frac{1}{\sqrt{x}} > 2*\sqrt{x+1} $ 
This is true iif : $ (2*\sqrt{x} +\frac{1}{\sqrt{x}})^2 > (2*\sqrt{x+1})^2 $
You get : $ 2*\sqrt{x} +\frac{1}{\sqrt{x}} > 2*\sqrt{x+1} $ <=> $ 4x + \frac{1}{x} +4 > 4*(x+1) $ <=> $ \frac{1}{x} > 0 $ which is true, so the inequality is true as well :) .
Now for the induction step : 
$ 2\sqrt{n+1} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} $ 
=> $2\sqrt{n+1} - 2 +\frac{1}{\sqrt{n+1}} < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}}$
But we have from the first part of the answer : $2*\sqrt{n+1} +\frac{1}{\sqrt{n+1}} -2 > 2*\sqrt{n+1+1} -2 = 2*\sqrt{n+2} -2$
Hence you get: $ 2*\sqrt{n+2} -2 < 2\sqrt{n+1} - 2 +\frac{1}{\sqrt{n+1}} < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}}$ Induction holds :)
