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# Finding a Function’s Upper Bound (TOMATO Subjective 144)

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 $$x^2> 0$$)

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

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