Find the minimum of $\arccos a+\arcsin a+\arctan a$

Question

Find the minimum of

$$\arccos a+\arcsin a+\arctan a$$

I know that, the answer is $\frac {\pi}{4}$.

But, I don't know a solution. Please, explain the solution of problem to me in a very simple way. Because, my mathematics level is too low to compare with you. Can you explain me the answer in detail, please? Thank you in advance.

Answer 1

Let $y = \arccos a+\arcsin a+\arctan a $

$\arccos a+\arcsin a =\frac{\pi}{2}$

So, $y = \frac{\pi}{2} + \arctan a $

$a$ lies on a subset of $[-1,1]$

Therefore, $y$ will be minimum when $a = -1$

Hence $y = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} $

Edit: $a$ lies on a subset of $[-1,1]$ because the $\arccos$ or $\arcsin$ function can't take values more than $1$ or less than $-1$