Question: Prove that $$\int\_0^{\large\frac{\pi}{2}}\sqrt{\sin x}\:dx >\frac{\pi}{3}$$

My work: $$\int\_0^{\large\frac{\pi}{2}}\sqrt{\sin x}\:dx > \int\_0^{\large\frac{\pi}{2}}\sin x \:dx = 1.$$

Answer: 
One may use the [standard identity](https://math.stackexchange.com/questions/980551/how-to-show-sin-x-geq-frac2x-pi-x-in-0-frac-pi2)
$$
\sin x\ge \frac{2x}{\pi},\qquad x \in \left[0,\frac \pi2 \right],
$$ giving
 $$\int\_0^{\frac{\pi}{2}}\sqrt{\sin x}\:dx >\int\_0^{\frac{\pi}{2}}\sqrt{ \frac{2x}{\pi}}\:dx=\frac{\pi}{3}$$ as wanted.
For a greater accuracy,
$$ \int\_{0}^{\pi/2}\sqrt{\sin x}\,dx = \int\_{0}^{1}\sqrt{\frac{u}{1-u^2}}\,du\stackrel{\text{Beta}}{=}\frac{\sqrt{2\pi}^3}{\Gamma\left(\frac{1}{4}\right)^2}=\text{AGM}(1,\sqrt{2})$$
([particular values of the $\Gamma$ function](https://en.wikipedia.org/wiki/Particular_values_of_the_gamma_function)) and
$$\text{AGM}\left(1,\sqrt{2}\right)=\text{AGM}\left(\frac{1+\sqrt{2}}{2},\sqrt[4]{2}\right)\geq \sqrt[8]{\frac{1+\sqrt{2}}{8}}\geq 1.1981. $$