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 ?

By C-S $$\arcsin^2{x}+\arccos^2{x}=\frac{1}{2}(1^2+1^2)(\arcsin^2{x}+\arccos^2{x})\geq\frac{1}{2}(\arcsin{x}+\arccos{x})^2=\frac{\pi^2}{8}.$$
The equality occurs for $x=\frac{1}{\sqrt2}$, which says that we got a minimal value.
In another hand, for $x\geq0$ we obtain: $$\arcsin^2{x}+\arccos^2{x}\leq\left(\arcsin{x}+\arccos{x}\right)^2=\frac{\pi^2}{4}.$$
The equality occurs for $x=0$.
For $x<0$ let  $t=-\arcsin{x}$.
Thus, $0<t\leq\frac{\pi}{2}$, $\arccos{x}=\pi-\arccos(-x)=\pi-\left(\frac{\pi}{2}-\arcsin(-x)\right)=\frac{\pi}{2}+t$ and 
$$\arcsin^2{x}+\arccos^2{x}=t^2+\left(\frac{\pi}{2}+t\right)^2\leq\frac{5\pi^2}{4}$$ with equality for $x=-1$, which says that $\frac{5\pi^2}{4}$ is a maximal value.
