Let $F:[0,1] \rightarrow \mathbb{R} $ be continuous. Suppose $F$ is differentiable on (0,1), $F(0) = 0$ and $F(x)>0$ for all $x \in (0,1)$.

Question

Let $F:[0,1] \rightarrow \mathbb{R} $ be continuous. Suppose $F$ is differentiable on (0,1), $F(0) = 0$ and $F(x)>0$ for all $x \in (0,1)$. Prove that there exists $r,s \in (0,1)$ such that $r + s = 1$ and $8\frac{F'(r)}{F(r)} = 5\frac{F'(s)}{F(s)}$

It seems that this question requires the mean-value theorem. I considered a function G(x) = F(x)F(1-x) and proved the case $\frac{F'(r)}{F(r)} = \frac{F'(s)}{F(s)}$, but for the case involving two different constants. I have no clue. Really thanks for any help or advice.

Answer 1

Apply Rolle's theorem to $$g(x):=8\ln F(x)+5\ln F(1-x).$$ $\lim_0g=\lim_1g=-\infty$ hence $g'$ must vanish somewhere inbetween.