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. 
 A: 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 
$f’(x_0)<1+f^2(x_0)$.
A: You can also go a more direct way after recognizing the connection to the tangent function: Consider the function 
$$
g(x)=\arctan(f(x)),
$$
then by the properties of the inverse tangent function, the interval length bound and the mean value theorem there exists some $x_0\in(a,b)$ so that
$$
\frac{\pi}4\ge\frac{|g(b)-g(a)|}{b-a}=|g'(x_0)|=\frac{|f'(x_0)|}{1+f(x_0)^2}
$$
A: Clearly scaling the function vertically or horizontally does not affect the question, so assume $a=0, b>4$, and WLOG $f(a)=0$. Let $y=f(x)$ and suppose for contradiction that $\frac{dy}{dx} \ge 1+y^2$ for all $x$ in the interval. Then we have $\frac{dx}{dy} \le \frac{1}{1+y^2}$, and integrating from gives $x \le \tan^{-1}{y}+c$. So $y>\tan{x-c}$. But $x$ varies over an interval of width at least $4\ge \pi$, so there exists some point where $\tan{x-c}$ approaches infinity, and then $y$, being continuous, must also approach infinity and become undefined at the asymptote. This is a contradiction, as desired.
