Set of values of $a$ satisfying an inverse trigonometric equation. 
Find the set of values of a so that the equation $(\arcsin x)^3+ (\arccos x)^3= a\pi^3$ has a solution.

Attempt: 
Let $y = \arcsin x$  (just for convenience to avoid typing/ writing $\arcsin$ over and over again )
Note that $y + \arccos x = \pi /2$
We have 
$(y+\arccos x)(y^2 + (\frac\pi 2-y)^2 - y (\frac \pi2 - y)= a\pi^3$
$\implies (\sqrt 3y- \dfrac{\sqrt3}{4}\pi)^2+\dfrac{\pi^2}{16}= 2a\pi^2$
$-\frac \pi 2\le y \le \frac\pi 2$
$\implies  -\dfrac{\sqrt 3 \pi}{2} - \sqrt{3}\pi/ 4\le \sqrt 3 y - \frac{\sqrt {3}}{4}\pi \le \dfrac{\sqrt 3 \pi}{2}-\sqrt{3}\pi/ 4\\ \implies 0\le 2a\pi^2- \dfrac{\pi^2}{16}\le \dfrac{3}{16}\pi^2 \implies a\in [\dfrac{1}{32}, \dfrac{1}{16}]$
I get the right lower limit of $a$ but the upper limit given in the answer is $7/8$. 
I'll be grateful if someone could point out my error and if that has been done, provide alternative, easier ways to solve this problem. 

The question is equivalent to finding the range of the function $$x\mapsto \frac{(\arcsin x)^3+ (\arccos x)^3}{\pi^3}$$
defined on $[-1,1]$.
You can do that by studying the variations of the function via its derivative, and the data of extremal values in $x=-1$ and $x=1$.

Edit
The problem is here:
$ -\dfrac{\sqrt 3 \pi}{2} - \sqrt{3}\pi/ 4\le \sqrt 3 y - \frac{\sqrt {3}}{4}\pi \le \dfrac{\sqrt 3 \pi}{2}-\sqrt{3}\pi/ 4$
from here, the largest value for the square will come from squaring the LHS, not the RHS.
