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 ?

Given, differentiate $f(x)$ with respect to $x$ now, $f'(x) = 0$ then, We also know, domain of  and  $∈ [ -1 1] ∴$ find $f(x)$ at $x = 1$ , $-1$ and $\frac{1}{\sqrt{2}}$ 
$f(1) = (sin^{-1}(1))^2 + (cos^{-1}(1))^2 = \frac{π^2}{4} + 0 = \frac{π^2}{4}$ $f(-1) = (sin^{-1}(-1))^2 + (cos^{-1}(-1))^2 = \frac{π^2}{4} + π^2 = \frac{5π^2}{4}$ $f(\frac{1}{\sqrt{2}}) = (sin^{-1}(\frac{1}{\sqrt{2}}))^{2} + (cos^{-1}(\frac{1}{\sqrt{2}}))^2 = \frac{π^2}{16} + \frac{π^2}{16} = \frac{π^2}{8}$
Hence, maximum value of $f(x) = \frac{5π^2}{4}$ Minimum value of $f(x) = \frac{π^2}{8}$
