Given $ a_1 =1 $ and $ a_{n+1}a_n=n+1 $. Prove that: $ \sum_{k=1}^n\frac{1}{a_k}\ge 2(\sqrt{n+1}-1) $.
Since $$\sqrt{n+1}-1 = \sum_{k=1}^n(\sqrt{k+1}-\sqrt{k}), $$ I attempt to prove that $$ 2a_n\leq \frac{1}{\sqrt{n+1}-\sqrt{n}}=\sqrt{n+1}+\sqrt{n}. $$ But it doesn't hold for some even $n$ (easy to find that $a_n=\frac{n!!}{(n-1)!!}$ and test with $ n = 2, 4, \cdots $).
Another way is to show that $$ \frac{1}{a_n}+\frac{1}{a_{n+1}} \geq 2(\sqrt{n+2}-\sqrt{n}). $$
This seems to hold with some $ n $ tested, but how to prove that? Or maybe the result doesn't hold at all.