TITLE: Prove inequality $\sqrt{\frac{1}n}-\sqrt{\frac{2}n}+\sqrt{\frac{3}n}-\cdots+\sqrt{\frac{4n-3}n}-\sqrt{\frac{4n-2}n}+\sqrt{\frac{4n-1}n}>1$
QUESTION [7 upvotes]: For any $n\ge2, n \in \mathbb N$ prove that
  $$\sqrt{\frac{1}n}-\sqrt{\frac{2}n}+\sqrt{\frac{3}n}-\cdots+\sqrt{\frac{4n-3}n}-\sqrt{\frac{4n-2}n}+\sqrt{\frac{4n-1}n}>1$$

My work so far:
1) $$\sqrt{n+1}-\sqrt{n}>\frac1{2\sqrt{n+0.5}}$$
2) $$\sqrt{n+1}-\sqrt{n}<\frac1{2\sqrt{n+0.375}}$$

REPLY [3 votes]: Another proof is this:
Note that 
$$
2 = \sqrt{\frac{4n}{n}} = \frac{1}{\sqrt{n}}\sum_{j=0}^{4n-1}\sqrt{j+1}-\sqrt{j}
$$ 
where the RHS can be expressed as 
$$
\frac{1}{\sqrt{n}}\left(\sum_{j=1}^{2n}(\sqrt{2j}-\sqrt{2j-1})+\sum_{j=1}^{2n-1}(\sqrt{2j-1}-\sqrt{2j-2})\right)
$$
Using that the function $f:(0,+\infty)\to \mathbb{R}$ given by $f(x)=\sqrt{x+1}-\sqrt{x}$ is strictly decreasing, we deduce, for all $j\in \{1,...,2n\}$, 
$$
\sqrt{2j-1}-\sqrt{2j-2}>\sqrt{2j}-\sqrt{2j-1}
$$
hence 
$$
2\sum_{j=1}^{2n}\left(\sqrt{\frac{2j-1}{n}}-\sqrt{\frac{2j-2}{n}}\right)>2
$$