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.