TITLE: Show $\frac{f'(x)}{f(x)}\ge\frac{g'(x)}{g(x)}-\left|\frac{f'(1)}{f(1)}-\frac{g'(1)}{g(1)} \right|$ if $\frac{f''(x)}{f(x)}\ge\frac{g''(x)}{g(x)}$.
QUESTION [6 upvotes]: Suppose $f$ and $g$ are positive increasing functions on $\left[ 1,+\infty \right)$ such that $$\frac{f''\left( x \right)}{f\left( x \right)}\ge \frac{g''\left( x \right)}{g\left( x \right)}\,.$$
Prove that
$$\frac{f'\left( x \right)}{f\left( x \right)}\ge \frac{g'\left( x \right)}{g\left( x \right)}-\left| \frac{f'\left( 1 \right)}{f\left( 1 \right)}-\frac{g'\left( 1 \right)}{g\left( 1 \right)} \right|
\,.$$

The problem is  taken from： Mitrinovic, D.S. Pecaric, j. E, Fink, A. M., Classical and new inequalities in analysis, Kluwer Acad tools. Publ, Dordrecht, 1993.
But I couldn't find the book, and I didn't come up with a good solution.
$$\varphi \left( x \right) =f'\left( x \right) g\left( x \right) -f\left( x \right) g'\left( x \right) \,\,$$
$$\varphi '\left( x \right) =f''\left( x \right) g\left( x \right) -f\left( x \right) g''\left( x \right) \ge 0$$
$$\varphi \left( x \right) \ge \varphi \left( 1 \right) \Rightarrow \frac{f'\left( x \right)}{f\left( x \right)}-\frac{g'\left( x \right)}{g\left( x \right)}\ge \frac{f\left( 1 \right) g\left( 1 \right)}{f\left( x \right) g\left( x \right)}\left[ \frac{f'\left( 1 \right)}{f\left( 1 \right)}-\frac{g\left( 1 \right)}{g\left( 1 \right)} \right] $$
The rest of the work is just to take the absolute value out of the case.

REPLY [4 votes]: You are almost done by proving that
$$
\frac{f'(x)}{f(x)}-\frac{g'(x)}{g(x)} \ge
\frac{f(1) g(1)}{f(x) g(x)} \left( \frac{f'(1)}{f(1)}-\frac{g(1)}{g(1)} \right)
$$
for $x \ge 1$. The right-hand side is
$$
 \ge - \frac{f(1) g(1)}{f(x) g(x)} \left| \frac{f'(1)}{f(1)}-\frac{g(1)}{g(1)} \right| \ge  - \left| \frac{f'(1)}{f(1)}-\frac{g(1)}{g(1)} \right|
$$
because $0 < \frac{f(1) g(1)}{f(x) g(x)} \le 1$.