Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
Hint: $f(x)-f(1)=\int_1^x f'(t)\,dt\le\int_1^x{1\over t^2+1}\,dt$. – David Mitra Jan 23 '13 at 13:39

1 Answer 1

up vote 4 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.