<s>[INST] Find the minimum value of a trigonometric function If the minimum value of $f\left(x\right)=\left(1+\frac{1}{\sin ^6\left(x\right)}\right)\left(1+\frac{1}{\cos ^6\left(x\right)}\right),\:x\:∈\:\left(0,\:\frac{\pi }{2}\right)$ is $m$, find $\sqrt m$.
How do I differentiate this function without making the problem unnecessarily complicated? If there are any other methods to finding the minimum value I am open to those too.
 [/INST] Without calculus, let $\,a = \sin^2(x)\,$, $\,b = \cos^2(x)\,$, then $\,f(x) = \left(1+\dfrac{1}{a^3}\right)\left(1+\dfrac{1}{b^3}\right)\,$, where $\,a,b \ge 0\,$ and $\,a+b=1\,$. By the generalized means inequalities:

*

*$\sqrt{ab} \le \dfrac{a+b}{2} = \dfrac{1}{2} \quad\implies\quad \dfrac{1}{ab} \ge 4 \quad\implies\quad \dfrac{1}{a^3b^3} \ge 4^3 = 64\;$;


*$\sqrt[3]{\dfrac{2}{\frac{1}{a^3}+ \frac{1}{b^3}}} \le \dfrac{a+b}{2} = \dfrac{1}{2} \quad\implies\quad \dfrac{1}{a^3}+ \dfrac{1}{b^3} \ge 16\;$.
Then the following inequality holds, with equality iff $a=b = \dfrac{1}{2}$, or $|\sin(x)| = |\cos(x)| = \dfrac{1}{\sqrt{2}}$:
$$
f(x) = \left(1+\frac{1}{a^3}\right)\left(1+\frac{1}{b^3}\right) \;=\; 1 + \frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{a^3b^3} \;\ge\; 1 + 16 + 64 = 81
$$
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