# Prove the following inequality: $\int_{0}^{\frac{\pi }{2}}\frac{sin(x)}{\sqrt{9-sin^{4}(x)}}dx\geq \frac{1}{3}$

Prove the following inequality: $$\int_{0}^{\frac{\pi}{2}}\frac{\sin x}{\sqrt{9-\sin^{4}x}}\ \mathrm dx\geq\frac{1}{3}.$$

I am thinking of replacing the equation with $$\int_{0}^{\frac{\pi }{2}}\frac{\sin x}{9-\sin^{4} x }dx\geq \frac{1}{3}$$, however I am stuck at this point.

Do you have any suggestions?

• use the fact $\sqrt{9 - \sin^4x} \leq 3$ and $\int_0^{\pi/2} sin x dx = 1$ and $\sin x$ is nonegative on $[0,\pi/2]$. Feb 5 '20 at 18:07
• You can get the proper font and spacing for $\sin$ using \sin. For operators that don't have a command of their own, you can use \operatorname{name}. Feb 5 '20 at 18:19

\begin{align*}\int_{0}^{\pi/2} \frac{\sin{x}}{\sqrt{9-\sin^4{x}}}dx &\geq \int_{0}^{\pi/2}\frac{\sin{x}}{\sqrt{9}}dx\\& = \frac{1}{3}\int_{0}^{\pi/2} \sin{x}dx \\&=\frac{1}{3}[-\cos{\pi/2}-(-\cos{0})]\\&=\frac{1}{3}[0+1]\\&=\frac{1}{3} \end{align*}
The inequality is because $$\sqrt{9-\sin^4{x}}\leq \sqrt{9}$$ due to positivity of $$\sin^4{x}$$. So ... $$\frac{1}{\sqrt{9-\sin^4{x}}}\geq \frac{1}{\sqrt{9}}$$
• It's also $< 1/(2\sqrt{2})$. Feb 5 '20 at 20:23
The given inequality can be deduced from AM-GM. Indeed$$\begin{eqnarray*} I=\int_{0}^{\pi/2}\frac{\sin x}{\sqrt{9-\sin^4 x}}\,dx &=& \int_{0}^{1}\frac{z\,dz}{\sqrt{(9-z^4)(1-z^2)}}\\ &=& \frac{1}{2}\int_{0}^{1}\frac{du}{\sqrt{(3-u)(3+u)(1-u)}}=\int_{0}^{1}\frac{dv}{\sqrt{(2+v^2)(4-v^2)}}\end{eqnarray*}$$ and over $$[0,1]$$ we have $$\sqrt{(2+v^2)(4-v^2)}=\text{GM}(2-v^2,4+v^2)\leq\text{AM}(2-v^2,4+v^2)=3$$.
This also shows that $$I$$ is an incomplete elliptic integral of the first kind.