How to prove the Theorem 148 in Inequalities by G. H. Hardy, J. E. Littlewood, G. Pólya?

Question

I am currently studying Inequalities by G. H. Hardy, J. E. Littlewood, G. Pólya, but I can not understand the proof about Theorem 148. Why can we know that there is only one intersection through the tangent line?

Answer 1

Since $f,g, \frac{f'}{g'}$ are all positive and increasing functions, their derivatives (if they exist) are positive. Since $g'/g$ is always strictly positive, we only have to look at $(f'/g' - f/g)$. There are 3 cases.

1. $\frac{f'}{g'} - \frac{f}{g} > 0$: $f/g$ is always increasing.

2. $\frac{f'}{g'} - \frac{f}{g} < 0$: $f/g$ is always increasing.

3. $\exists x_0$ such that $\frac{f'(x_0)}{g'(x_0)} - \frac{f(x_0)}{g(x_0)} =0.$ At such an $x_0$, $\frac{d}{dx}(f/g) = 0$. However, we know that $f'/g'$ is increasing (so it either has a positive jump or a positive derivative), and $f/g$ has a zero derivative at that point, so locally around $x_0$, $$\frac{f'}{g'} - \frac{f}{g} \text{ is strictly increasing.}$$

(I think this is what is being expressed by the tangent lines.)

In terms of a "sign test" for the derivative, $\frac{d}{dx}(f/g)$ would be negative to the left of $x_0$ and positive to the right of $x_0$.

If there was a second point $x_1 > x_0$ such that $\frac{d}{dx}(f/g)(x_1) = 0$, this would contradict the fact that the sign of our derivative to the left of $x_1$ would be negative.