Prove that $\frac{\pi}{4}\le\sum_{n=1}^{\infty} \arcsin\left(\frac{\sqrt{n+1}-\sqrt{n}}{n+1}\right)$

Prove that $$\frac{\pi}{4}\le\sum_{n=1}^{\infty} \arcsin\left(\frac{\sqrt{n+1}-\sqrt{n}}{n+1}\right)$$

EDIT: inspired by Michael Hardy's suggestion I got that

$$\arcsin \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{(n+1)(n+2)}}=\arcsin\frac{1}{\sqrt{n+1}}-\arcsin\frac{1}{\sqrt{n+2}}$$ and then $$\sum_{n=1}^{\infty} \arcsin\left(\frac{\sqrt{n+1}-\sqrt{n}}{n+1}\right)\ge\sum_{n=1}^{\infty} \arcsin\left(\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{(n+1)(n+2)}}\right)\rightarrow\frac{\pi}{4}$$ because $\sum_{n=1}^{\infty} \left(\arcsin\frac{1}{\sqrt{n+1}}-\arcsin\frac{1}{\sqrt{n+2}}\right)=\arcsin \frac{\sqrt{2}}{2}=\frac{\pi}{4}$

Sis & Chris.

Since $\arcsin x\ge \arctan x$ for $x \in [0,1]$, thus we shall have $$\arcsin(\frac{\sqrt{n+1}-\sqrt{n}}{n+1})\ge \arctan(\frac{\sqrt{n+1}-\sqrt{n}}{n+1})\ge \arctan\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}\sqrt{n}+1} $$ $$= \arctan{\sqrt{n+1}}-\arctan{\sqrt{n}}.$$

(The last equality uses $\arctan(x) - \arctan(y) = \arctan(\frac{x-y}{1 + xy})$.)

Done.

P.S.

$\sum \arcsin(\frac{\sqrt{n+1}-\sqrt{n}}{n+1}) \ge \sum(\arctan{\sqrt{n+1}}-\arctan{\sqrt{n}}) = \pi/2-\arctan(1)=\pi/4$