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I came across this question while learning about definite integrals.

$$\int_0^{\pi/2} \sqrt{\sin x} dx \lt \sqrt\frac{\pi}2$$

Understanding that I cannot calculate the actual value of the integral of $\sqrt{\sin x}$ I tried the following: First, I tried to prove that since sin x is an increasing function over 0 to $\pi/2$, if:

$$\sqrt{\frac{2}{\pi}} \gt \sqrt {\sin x}$$

for x in 0 to $\frac{\pi}{2}$ then the given inequality would hold true but clearly for certain values this is not true. Then, I tried creating an integral like:

$$\int_0^{\pi/2} \frac{1}{2\sqrt{x}} dx = \sqrt\frac{\pi}2$$

so if we can prove that $\frac{1}{2\sqrt{x}} \gt \sqrt{\sin x}$ for all x in 0 to $\frac{\pi}{2}$ this would also prove the given inequality but 1 is not greater than $4x \sin x$ for all values of x in this range.

Svee
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    Apply Cauchy Schwarz inequality (the version for integrals). If you haven't learn that yet, look at integral $\int_0^{\frac{\pi}{2}}\left(\sqrt{\sin x}-\sqrt{\frac2{\pi}}\right)^2dx$. – achille hui Oct 27 '21 at 01:17
  • @achillehui unfortunately I haven't learnt that yet. What am I supposed to see in the integral you mentioned? – Svee Oct 27 '21 at 01:23
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    The integral $> 0$. if you expand the integrand and integrate the terms you know, you will get the inequality you need. – achille hui Oct 27 '21 at 01:26
  • Oh that's a really slick solution. How did you come to think of it? – Svee Oct 27 '21 at 01:32
  • This is essentially one of the way to prove CS inequality. Given $(a_k), (b_k) \in \mathbb{R}^n$, not identically zero. Let $\lambda = \sqrt{\frac{\sum_{k=1}^n a_k^2}{\sum_{k=1}^n b_k^2}}$, then $\sum\limits_{k=1}^n(a_k - \lambda b_k)^2 \ge 0$. If you expand the summand, you get the CS inequality. For the integral version, you just replace the sums by integrals. – achille hui Oct 27 '21 at 01:40
  • Oh I see the inequality seems interesting. Will study up on it. Thanks! – Svee Oct 27 '21 at 01:45

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