An integral inequality

If $f$ is continuous and its derivative is strictly decreasing in $[0,1]$ then prove that

$\int_{0}^{1} \frac{\mathrm{d}x}{1+f^2(x)} \leq \frac{f(1)}{f'(1)}$

if it is also known that $f(0)=0$ and $f'(1)>0$ .

Solution

Since $f'$ is strictly decreasing then $f'(x) \geq f'(1)>0$ . Thus, $f$ is strictly increasing and $f(x)>0$ . Therefore,

$\begin{align*} \int_{0}^{1} \frac{f'(1)}{1+f^2(x)}\,\mathrm{d}x &\leq \int_{0}^{1} \frac{f'(x)}{1+f^2(x)}\, \mathrm{d}x \\ &= \left [ \arctan f(x) \right ]_0^1 \\ &= \arctan f(1) \\ &\leq f(1) \end{align*}$

since $\arctan x \leq x$ for all $x>0$ .

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