<s>[INST] Minimum value of inverse trigo function (cubic) The minimum value of $(\sin^{-1}x)^3+(\cos^{-1}x)^3$ is equal to ( following options)
a) $\displaystyle \frac{\pi^3}{32}$
b) $\displaystyle\frac{5\pi^3}{32}$
c) $\displaystyle\frac{9\pi^3}{32}$
d) $\displaystyle\frac{11\pi^3}{32}$
Can we go like this : 
$ -\frac{\pi}{2} \leq \sin^{-1}x \leq \frac{\pi}{2}$  Therefore minimum value of $(\sin^{-1}x)^3 = (\frac{-\pi}{2})^3$= -$\frac{\pi}{8}$
Please guide..
 [/INST] Using $$\frac{(\arcsin x)^3+(\arccos x)^3}{2}\geq \bigg(\frac{\arcsin x+\arccos x}{2}\bigg)^3$$
So $$\displaystyle (\arcsin x)^3+(\arccos x)^3\geq \frac{1}{4}\cdot \frac{\pi^3}{8} = \frac{\pi^3}{32}$$
and equality hold when $\displaystyle \arcsin x = \arccos x\Rightarrow x = \frac{1}{\sqrt{2}}$
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