Deriving and proving the upper bound for a function when $x$ tends to $\infty$

Question

Let $f(x)$ be a real valued function defined on the interval $[1,\infty)$. The following equations hold:
$$f(1)=1$$ $$f'(x)=\dfrac{1}{x^2+(f(x))^2}$$ Prove that $\lim_{x\to \infty}f(x)<1+\dfrac{\pi}{4}$

I have no idea how to proceed with this question. Please help.

Thank You

Edit: Working on the hint given by Atmos, it goes as follows:

Observe that $f'(x)>0$. This means that $f(x)$ is a strictly increasing function.

So $f(x)\geq 1\space \forall\ x\in [1,\infty)\Rightarrow x^2+(f(x))^2\geq x^2+1\space \forall\ x\in [1,\infty)$

Thus $$f'(x)=\dfrac{1}{x^2+(f(x))^2}\leq \dfrac{1}{x^2+1}$$

Since inequality is preserved while integration, integrating both sides give: $$f(x)\leq \arctan(x)+C$$

now we have $f(1)=1\Rightarrow C=1-\dfrac{\pi}{4}$

Also $f(x)\leq \arctan(x)+1-\dfrac{\pi}{4}\leq \dfrac{\pi}{2}+1-\dfrac{\pi}{4}=1+\dfrac{\pi}{4}$

Hence we are done.

Answer 1

Hint :

What can you say about the variation of $f$. What lower bound for $f$ can you deduce from this ?

Answering this question, try to show that $$ f'\left(x\right) \leq \frac{1}{1+x^2} $$

I let you conclude.