Proving that $b-a\ge \pi $ Let $f : C^1(a; b)$, such that $ \lim\limits_{x\to a^+}
f(x) = +\infty$,  $\lim\limits_{x\to b^-}f(x) = -\infty$ and 
$f'(x)+f^2(x) \ge -1 $ for $x \in  (a; b)$. 

Prove that $b-a\ge \pi $ and provide an example
  where $b-a= \pi $

For the second question the obvious example could be $f(x) = \cot(x)$ with $a=0$, and $b=\pi.$ 
Any hint for the first part?
This queston is similar to this: How prove this inequality $b-a\ge \pi$

Since $f^{2}+1$ is never zero, the inequality transforms into
$$\frac{f'(x)}{1+f^{2}(x)}\geq -1$$
We integrate this inequality from $a$ to $b$, and the LHS becomes 
\begin{align}
\int_{a}^{b}\frac{f'(x)}{1+f^{2}(x)}\text{d} &= \int_{a}^{b}(\arctan{f(x)})'\text{d} x\\
&=-\frac{\pi}{2} -( \frac{\pi}{2})\\
&=-\pi
\end{align}
The RHS of the inequality is just $-(b-a)$, therefore
$$-\pi\geq -(b-a)\Rightarrow b-a\geq \pi$$ 
