Find the greatest and least values of $(\sin^{-1}x)^2+(\cos^{-1}x)^2$ Find the upper and lower limit of 

$$
(\sin^{-1}x)^2+(\cos^{-1}x)^2
$$

My Attempt:
$$
\frac{-\pi}{2}\leq\sin^{-1}x\leq \frac{\pi}{2}\quad\&\quad0\leq\cos^{-1}x\leq\pi\\(\sin^{-1}x)^2\leq\frac{\pi^2}{4}\quad\&\quad(\cos^{-1}x)^2\leq\pi^2\\
0\leq(\sin^{-1}x)^2+(\cos^{-1}x)^2\leq\frac{\pi^2}{4}+\pi^2=\frac{5\pi^2}{4}
$$
Here, I can see the upper limit is $\frac{5\pi^2}{4}$ which is fine. But, $0$ is one lower limit not the lower limit.
Why am I not getting the lower limit in my approach ?
How do I approach similar problems involving max and min, when you don't get the lower or upper limits ?

You can do it with calculus: if $f(x)=(\arcsin x)^2+(\arccos x)^2$, then
$$
f'(x)=\frac{2\arcsin x}{\sqrt{1-x^2}}-\frac{2\arccos x}{\sqrt{1-x^2}}
$$
(for $-1<x<1$). The derivative vanishes for $\arcsin x=\arccos x$, that is, $x=1/\sqrt{2}$. Since
$$
f(1/\sqrt{2})=\frac{\pi^2}{8}
\qquad
f(-1)=\frac{5\pi^2}{4}
\qquad
f(1)=\frac{\pi^2}{4}
$$
you can easily draw the conclusion.
