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.