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?

  • 4
    $\begingroup$ use the fact $\sqrt{9 - \sin^4x} \leq 3$ and $\int_0^{\pi/2} sin x dx = 1$ and $\sin x$ is nonegative on $[0,\pi/2]$. $\endgroup$
    – Chival
    Feb 5, 2020 at 18:07
  • 1
    $\begingroup$ 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}. $\endgroup$
    – joriki
    Feb 5, 2020 at 18:19

2 Answers 2


\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 ... $$\frac{1}{\sqrt{9-\sin^4{x}}}\geq \frac{1}{\sqrt{9}}$$

  • $\begingroup$ It's also $< 1/(2\sqrt{2})$. $\endgroup$ Feb 5, 2020 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.


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