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?

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