Prove that $$\int_0^{\large\frac{\pi}{2}}\sqrt{\sin x}\:dx >\frac{\pi}{3}$$
My work: $$\int_0^{\large\frac{\pi}{2}}\sqrt{\sin x}\:dx > \int_0^{\large\frac{\pi}{2}}\sin x \:dx = 1.$$
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One may use the standard identity $$ \sin x\ge \frac{2x}{\pi},\qquad x \in \left[0,\frac \pi2 \right], $$ 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.
Olivier Oloa
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Nice answer (+1)! Wouldn't it be better to say that $x\in (0,\pi/2)$ in the first line so that you have a strict inequality? That justifies the second strict inequality. – Shashi Jan 14 '18 at 22:00
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I believe the inequality being used is known as Jordan's inequality. – omegadot Jan 14 '18 at 22:47
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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) and $$\text{AGM}\left(1,\sqrt{2}\right)=\text{AGM}\left(\frac{1+\sqrt{2}}{2},\sqrt[4]{2}\right)\geq \sqrt[8]{\frac{1+\sqrt{2}}{8}}\geq 1.1981. $$
Jack D'Aurizio
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