If $f:[0,1] \rightarrow[0, \infty)$ is differentiable function with decreasing first derivative such that $f(0)=0$ and $f'(x)>0$, then

$\int\limits_0^1 \frac{1}{f^2(x)+1} d x \leq \frac{\tan ^{-1}(f(1))}{f'(1)}$

It is given that $f'(x)$ is decreasing on $[0,1]$.

Therefore, $f'(1)$ is the least value of $f'(x)$ in $[0,1]$

∴   $f'(x) \geq f'(1)$ for all $x \in[0,1]$

$\Rightarrow \frac{f'(x)}{f^2(x)+1} \geq \frac{f'(1)}{f^2(x)+1}$

$\Rightarrow \int\limits_0^1 \frac{f'(x)}{f^2(x)+1} d x \geq \int\limits_0^1 \frac{f'(1)}{f^2(x)+1} d x$

$\Rightarrow \tan ^{-1}(f(1))-\tan ^{-1}(f(0)) \geq \int\limits_0^1 \frac{f'(1)}{f^2(x)+1} d x$

$\Rightarrow \tan ^{-1}(f(1)) \geq \int\limits_0^1 \frac{f'(1)}{f^2(x)+1} d x$

$\Rightarrow \left[\tan ^{-1}(f(x))\right]_0^1 \geq \int\limits_0^1 \frac{f'(1)}{f^2(x)+1} d x$