I have an inequality as shown below
$$\int_0^{\frac{\pi}{2}}\sqrt{1+a\sin^2x} \geq \frac{\pi}{4}(1+\sqrt{1+a})\ \ \ \ \ \ \ (a>-1)$$
I want to know how to use the conventional calculus method to prove it.
How can I get it? Any help will be appreciated.
Proof calculus inequality $\int_0^{\frac{\pi}{2}}\sqrt{1+a\sin^2x} \geq \frac{\pi}{4}(1+\sqrt{1+a})$
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$\begingroup$ I've seen this (or similar) question before. The $\int_0^{\pi/4}$ part should be $\int_0^{\pi/2}$. I plotted both in Mathematica and it showed that the former one is incorrect. $\endgroup$– Kemono ChenCommented Aug 31, 2019 at 3:07
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$\begingroup$ Take $a=0$, you will see what? $\endgroup$– RiemannCommented Aug 31, 2019 at 3:08
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$\begingroup$ yes,it shuold be $\int_{0}^{\pi/2}$ $\endgroup$– XinCommented Aug 31, 2019 at 3:17
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1 Answer
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$$\int_0^{\pi/2}\sqrt{1+a\sin^2x}dx=\int_0^{\pi/2}\sqrt{\cos^2x+(1+a)\sin^2x}\sqrt{\sin^2x+\cos^2x}dx\\ \ge\int_0^{\pi/2}(\cos^2 x+\sqrt{1+a}\sin^2 x)dx\\ =\frac\pi4(1+\sqrt{1+a}) $$ Explanation: the non-trivial step is $(a_1b_1+a_2b_2)^2\le(a_1^2+a_2^2)(b_1^2+b_2^2)$.
answered Aug 31, 2019 at 3:27