Question

Suppose $f$ is a real-valued differentiable function defined on $[1,\infty )$ with $f\left(1\right)=1.$ Moreover, suppose that $f$ satisfies$f^{\prime }\left(x\right)=\frac{1}{x^2+f^2\left(x\right)}.$ Show that $f\left(x\right)<1+\frac{\pi }{4}\forall x\ge 1$

Answer 1

Here $,f^{\prime }\left(x\right)=\frac{1}{x^2+\left(f\right(x){)}^2}>0\forall x\ge 1$

$\Rightarrow f\left(x\right)$ is an increasing function $\forall x\ge 1$

Given

$f\left(1\right)=1\Rightarrow f\left(x\right)\ge 1\forall x\ge 1$

Hence, $f^{\prime }\left(x\right)\le \frac{1}{1+x^2}\forall x\ge 1$

$\Rightarrow {\int }_1^x{f}^{\prime }\left(x\right)dx\le {\int }_1^x\frac{1}{1+x^2}dx$

$\Rightarrow f\left(x\right)-f\left(1\right)\le {\tan }^{-1}x-{\tan }^{-1}1$

$\Rightarrow f\left(x\right)\le {\tan }^{-1}x+1-\frac{\pi }{4}$

$\Rightarrow f\left(x\right)<\frac{\pi }{2}+1-\frac{\pi }{4}$

$(\text{as}{\tan }^{-1}x<\frac{\pi }{2},\forall x\ge 1)$

i.e., $f\left(x\right)<1+\frac{\pi }{4}\forall x\ge 1$