Inequality with sines and cosines

Let $0<\theta< \frac{\pi}{2}$. Prove that:

$$\sqrt{\sin^2 \theta +\frac{1}{\sin^2 \theta}}+ \sqrt{\cos^2 \theta + \frac{1}{\cos^2 \theta}}\geq \sqrt{10}$$

Solution (George Basdekis)



We set $x= \sin^2 \theta, \; y=\cos^2 \theta$ and we note that $x+y=1$. Hence we need to prove that:

$$\sqrt{x+ \frac{1}{x}}+ \sqrt{y+ \frac{1}{y}}\geq \sqrt{10}, \;\;\;\; x+y=1$$

From the AM-GM & Cauchy Schwartz inequality we have that:

$$\sqrt{x+ \frac{1}{x}}+ \sqrt{y+ \frac{1}{y}}\geq 2\sqrt[4]{\left ( x+ \frac{1}{x} \right )\left ( y+ \frac{1}{y} \right )}\geq 2 \sqrt{\sqrt{xy}+ \frac{1}{\sqrt{xy}}}$$

So, it suffices to prove that $\displaystyle \sqrt{xy}+ \frac{1}{\sqrt{xy}}\geq \frac{5}{2}$.  From the AM- GM inequality we see that:

$$\begin{aligned}\sqrt{xy}+\frac{1}{\sqrt{xy}}=\sqrt{xy}+4\cdot\frac{1}{4\sqrt{xy}}&\geq 5\sqrt[5]{\sqrt{xy}\cdot\frac{1}{4^4(\sqrt{xy})^4}}\\&=5\sqrt[5]{\frac{1}{2^8}\cdot\frac{1}{(\sqrt{xy})^3}}\\&\geq 5\sqrt[5]{\frac{1}{2^8}\cdot\frac{1}{(\frac{x+y}{2})^3}}\\&=5\sqrt[5]{\frac{1}{2^8}\cdot 2^3}\\&=\frac{5}{2}\end{aligned}$$

completing the exercise.