For any $n\ge2, n \in \mathbb N$ prove that $$\sqrt{\frac{1}n}-\sqrt{\frac{2}n}+\sqrt{\frac{3}n}-\cdots+\sqrt{\frac{4n-3}n}-\sqrt{\frac{4n-2}n}+\sqrt{\frac{4n-1}n}>1$$
1) $$\sqrt{n+1}-\sqrt{n}>\frac1{2\sqrt{n+0.5}}$$
2) $$\sqrt{n+1}-\sqrt{n}<\frac1{2\sqrt{n+0.375}}$$
We have to prove that:
$$ \sum_{k=1}^{2n}\sqrt{2k-1}-\sum_{k=1}^{2n}\sqrt{2k-2} > \sqrt{n} \tag{1}$$
hence it looks like a good idea to apply creative telescoping and approximate:
$$\sqrt{2k-1}-\sqrt{2k-2}\geq\frac{\sqrt{k-1/4}-\sqrt{k-5/4}}{\sqrt{2}}-\frac{1}{128\sqrt{2}}\left(\frac{1}{(k-5/4)^{3/2}}-\frac{1}{(k-1/4)^{3/2}}\right)\tag{2} $$
I found $(2)$ by playing a bit with the Laurent expansion of the LHS in a neighbourhood of $+\infty$.
It is an algebraic inequality not terribly difficult to prove once established, and the RHS is a telescopic term, so, by summing it over $k=2,\ldots,2n$, we get an inequality actually (slightly) stronger than the wanted one.
A simpler approach may be to show that $A_n$ defined through $$ A_n = \sum_{k=1}^{2n}\left(\sqrt{2k-1}-\sqrt{2k-2}\right) $$ fulfils $A_n^2 \geq 1+A_{n-1}^2$ by induction.
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 $$ 2\sum_{j=1}^{2n}\left(\sqrt{\frac{2j-1}{n}}-\sqrt{\frac{2j-2}{n}}\right)>2 $$
