# Problem

$ f(x)$ is defined over $ [a,b]$ and differentiable over $ (a,b)$ , where $ b-a\geq 4.$ Prove that there exists $ \xi \in (a,b)$ such that $ f'(\xi)<1+f^2(\xi)$ .

# My Proof

Since $ b-a \geq 4$ ，we can obtain $ $ \exists x_1,x_2 \in (a,b):x_2-x_1>\pi.$ $ Denote$ $ F(x):=\arctan f(x).$ $ Obviously, $ F(x)$ is continuous over $ [x_1,x_2]$ and differentiable over $ (x_1,x_2)$ . Thus, By Lagrange’s Mean Value Theorem，$ $ \exists \xi \in (\xi,x_2) \subset (a,b):F(x_2)-F(x_1)=F'(\xi)(x_2-x_1).$ $ Further, $ $ \frac{f'(\xi)}{1+f^2(\xi)}=F'(\xi)=\frac{F(x_2)-F(x_1)}{x_2-x_1}\leq \frac{|F(x_2)|+|F(x_1)|}{x_2-x_1}<\frac{\frac{\pi}{2}+\frac{\pi}{2}}{\pi}=1,$ $ Which implies $ $ f'(\xi)<1+f^2(\xi).$ $

**AM I RIGHT? HOPE TO SEE OTHER PROOFS. THX.**