Trying to prove that
$$ \sum_{i=1}^n \frac1{\sqrt i} > 2(\sqrt{n+1}-1), \forall n \in \mathbb N$$
using induction.
My only attempt so far has consisted of squaring both sides (during the $P_{k+1}$ part) to get rid of square roots, but it turned real messy, real fast.
Tried using
$$\Bigg[\sum_{i=1}^{k+1} \frac1{\sqrt i}\Bigg]^2 > \Bigg[ 2(\sqrt{k+1}-1) + \frac{1}{\sqrt{k+1}}\Bigg]^2$$
but the algebra was just too dirty.
Hint. How about using the inequality
$$ \frac{1}{\sqrt{i}} = \int_{i}^{i+1} \frac{dx}{\sqrt{i}} > \int_{i}^{i+1} \frac{dx}{\sqrt{x}} = [2\sqrt{x}]_{i}^{i+1} = 2\sqrt{i+1} - 2\sqrt{i}. $$
Hint. To make proof by induction work, you need a relationship (fairly simple if possible) between the statement for $k$ and the statement for $k+1$. In this case the relationship is that the LHS for $k+1$ is equal to the LHS for $k$ plus one more term. So you need to prove that $$2(\sqrt{k+1}-1)+\frac{1}{\sqrt{k+1}}>2(\sqrt{k+2}-1)\ .$$ If you do some working to eliminate the denominator, then eliminate the square roots, the algebra is not actually all that awful. Good luck!
Trick is: $\dfrac{1}{2\sqrt{i}} > \dfrac{1}{\sqrt{i+1} + \sqrt{i}} = \sqrt{i+1} - \sqrt{i}$. Thus:
$\displaystyle \sum_{i=1}^n \dfrac{1}{2\sqrt{i}} > \displaystyle \sum_{i=1}^n \left(\sqrt{i+1} - \sqrt{i}\right) = \sqrt{n+1} - 1$.
If $$\sum_{i=1}^n \frac{1}{\sqrt i} > 2(\sqrt{n+1}-1)$$ for some $n$, then $$\sum_{i=1}^{n+1} \frac{1}{\sqrt i} = \sum_{i=1}^{n} \frac{1}{\sqrt i} + \frac{1}{\sqrt{n+1}} > 2(\sqrt{n+1}-1) + \frac{1}{\sqrt{n+1}}$$ so it suffices to show that $$2(\sqrt{n+1}-1) + \frac{1}{\sqrt{n+1}} \geq 2(\sqrt{n+2}-1).$$ This is equivalent to showing $$\frac{1}{\sqrt{n+1}} \geq 2\sqrt{n+2} - 2\sqrt{n+1}$$ and in turn to $$\sqrt{\frac{n+2}{n+1}} + 1 \geq 2(n+2 - (n+1))= 2$$ (by multiplying both sides by $\sqrt{n+2} + \sqrt{n+1}$. This last one is easily seen to hold, as $\sqrt{\frac{n+2}{n+1}} > 1$.
