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 ?

$$(\arcsin x)^2+(\arccos x)^2=(\arccos x+\arcsin x)^2-2\arccos x\cdot \arccos x\leq \left(\frac{\pi}{2}\right)^2=\frac{\pi^2}{4}.$$
Equality hold when $x=0$ or $\displaystyle x=1$
Similarly using $$\frac{(\arcsin x)^2+(\arccos x)^2}{2}\geq \bigg(\frac{\arcsin x+\arccos x}{2}\bigg)^2$$
so $$(\arcsin x)^2+(\arccos x)^2\geq \frac{\pi^2}{8}$$
equality hold when $\displaystyle x=\frac{1}{\sqrt{2}}$.
And at $x=-1$ value of $\displaystyle (\arcsin x)^2+(\arccos x)^2=\frac{5\pi^2}{4}$
$$\Rightarrow  \frac{\pi^2}{8}\leq (\arcsin x)^2(\arccos x)^2\leq \frac{5\pi^2}{4}$$
