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Suppose $f$ is a real valued differentiable function defined on $[1,\infty)$ with $f(1)=1$. Suppose also that $f$ satisfies $$f'(x)=\frac{1}{x^2+f^2(x)}.$$ The question is to prove that $f(x) \leq 1+\pi/4$ for every $x \geq 1$

I tried to solve the differential equation but could not bring it in some known form. I examined the derivative of $\tan^{-1}x$ which looks similar to that in the question. However I could not get any idea with that. Any help shall be highly appreciated. Thanks.

Navin
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1 Answers1

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Since $f'(x) > 0$ for every $x\geq 1$, the function $f$ is (strictly) monotone increasing in $[1,+\infty)$, hence $f(x) > 1$ for every $x > 1$. As a consequence, $f'(x) = \frac{1}{x^2 + f^2(x)} \leq \frac{1}{x^2+1}$ for every $x\geq 1$. Moreover, $f$ is a $C^1$ function, hence for every $x\geq 1$ $$ f(x) = f(1) + \int_1^x f'(t)\, dt \leq 1 + \int_1^x \frac{1}{1+t^2}\, dt = 1 + \arctan x - \frac{\pi}{4} < 1 + \frac{\pi}{4}. $$

Rigel
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