TITLE: How to show $f(x) \leq 1+\frac{\pi}{4}$ for every $x \geq 1$
QUESTION [5 upvotes]: Suppose $f$ is a real-valued differentiable function  defined on $[ 1,\infty)$ with $f(1)=1$. Suppose , moreover , that $f$ satisfies 
   $$f'(x)=\frac{1}{x^2+f^2(x)}$$
  Show that $f(x) \leq 1+\frac{\pi}{4}$ for every $x \geq 1$ .

Trial: I try to find the maximum value of $f(x)$ but here $f'(x)=0$ has no solution.

REPLY [4 votes]: as david said:From the given condition  $f'(x)={1\over x^2+f^2(x)}$, it follows that   $f'$ is positive on $[1,\infty)$. So $f$ is increasing on $[1,\infty)$, and since $f(1)=1$, we have $f(x)\ge1$ on $[1,\infty)$.
The aforementioned condition also implies that $f'$ is continuous on $[1,\infty)$. We thus have 
$$
f(t)-f(1)=\int_1^t {1\over x^2+f^2(x)}\,dx\le \int_1^\infty{1\over 1+x^2}\,dx=\tan^{-1} x\,\bigl|_1^\infty={\pi\over 4};
$$
whence the result follows.