Prove that $2\arctan x + \arcsin \frac{2x}{1+x^2} = \pi$ for every $x\geq1$

Question

Prove that for every $x\geq1$ $$f(x) = 2\arctan x + \arcsin \frac{2x}{1+x^2} = \pi$$

My idea is to firstly calculate $f(1)$ which is actually $\pi$. Then I need to show, that for every $x\geq1$, derivative of $f(x)$ is equal to $0$

However, differentiation is a little bit complicated. This is output of WolframAlpha:

What is smarter method to prove this inequality?

Answer 1

You can simplify the expression in the formula of the derivative further:

$$\frac{\sqrt \frac{(x^2-1)^2}{(x^2+1)^2}}{x^2-1}$$

$$=\frac{ \frac{x^2-1}{x^2+1}}{x^2-1}$$

$$= \frac{1}{x^2+1}$$

So you get that the derivative is zero.