real analysis - Prove there exists some $x_0$ for a differentiable function

A real-valued function $f$ is defined and differentiable on $[a,b]$ ($b-a\geq{4}$).
Prove that there exists $x_0 \in (a,b)$ for which $f'(x_0)<1+f^2(x_0)$

On the one hand, the statement resembles very much the classical theorem by Lagrange according to which there exists some $\varepsilon$ for which $f'(\varepsilon)(b-a)=f(b)-f(a)$

Nevertheless, what we have here is an inequality, which is more tricky.
The limitation ($b-a\geq{4}$) makes me think that using trigonometry, in this case, might be a good approach, but I have no idea how exactly that could be used.

Thanks in advance for any hints.

Answer

If this is not true we would get
$$\frac{f’(x)}{1+f^2(x)}\ge 1$$ for all $x$ in $[a,b]$.
Integrating we get
$$\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)>\arctan(f(b))-\arctan(f(a))\ge b-a$$
That is $b-a<\pi<4$, so, if $b-a>4$ (or even $b-a\ge \pi$) then there must be some $x_0\in [a,b]$ such that
$f’(x_0)<1+f^2(x_0)$.