I am trying to prove the following hypothesis:
Let $F \in \mathcal{C}^1$ be a function such that $x(F-xF')$ has a determined sign, meaning that $x(F-xF')\geq 0$ $(or \leq 0)$ for all $x\in\mathbb{R}$ where the equality holds only for a set of Lebesgue measure $0$. If there exist $a,b \in \mathbb{R}$ such that $F(a)^2 - 4a^2 = F(b)^2 - 4b^2 = 0$, then for all $x \in <a,b>$, $F(x)^2 - 4x^2<0$.
So far I haven't been able to prove the statement nor find a counterexample so any help would be appreciated!