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Problem: Suppose $f(x)$ is a real valued differentiable function defined on $[1,\infty)$ with $f(1)=1$. Suppose moreover $f(x)$ satisfies

$f'(x) = \frac {1}{x^2+f^2(x)}$

Show that $f(x) \leq 1+\frac{\pi}{4}$ for every $x \geq 1$

Solution: As the question doesn’t requires us to find an exact solution rather just an upper bound, we can easily find it by manipulating the given statement after establishing certain properties of $f(x)$.

We see that $f'(x)>0$ for all $x$ which means $f(x)$ is an increasing function.

As the domain is $[1,\infty)$ we can say that $f(x)\geq f(1)$ for all $x$.

$=> f^2(x)\geq f^2(1)$

$=> x^2+f^2(x)\geq x^2+f^2(1)$         (as $x^2> 0$)

$=> \frac{1}{x^2+f^2(x)}\leq \frac{1}{x^2+f^2(1)}$

$=> f'(x)\leq \frac{1}{x^2+f^2(1)}$

Integrating both sides from 1 to $x$

$=> \int_{1}^{x}f'(x)\leq \int_{1}^{x}\frac{1}{x^2+f^2(1)}\leq \int_{1}^{\infty}\frac{1}{x^2+f^2(1)}$

As $f(1)=1$ we have,

$=> \int_{1}^{x}f'(x)\leq \int_{1}^{\infty}\frac{1}{x^2+1}$

$=> f(x) - f(1) \leq tan^{-1}\infty - tan^{-1}1$

$=> f(x) - f(1) \leq \frac{\pi}{2} - \frac{\pi}{4}$

Substituting $f(x) = 1$

$=> f(x) \leq 1+\frac{\pi}{4}$

Hence Proved.