Let $f(x)$ and $g(x)$, $x>0$ positive, smooth functions on $\mathbb{R}_+$. Assume that $f(x) > g(x)$ for all $x$, and that both $f(x)$ and $g(x)$ are strictly decreasing. Further, assume that $f'(x) < g'(x)$ for all $x$, i.e., $f(x)$ is decreasing faster than $g(x)$.
Is it then the case that the ratio
$$ \frac{f(x)}{g(x)} $$ is a decreasing function of $x$?
On
It might be late, but taking upon @Adayah's answer you need to show that
$$\frac{f'(x)}{f(x)}<\frac{g'(x)}{g(x)}$$
Thinking about $\frac{d}{dx}\log f(x)= \frac{f'(x)}{f(x)}$, we may express the above inequality as
$$\frac{d}{dx}\log f(x)<\frac{d}{dx}\log g(x).$$ When $f(x)$ and $g(x)$ are continuous and since the exponential is strictly increasing and continuous you may pull the exponential into the differential-operator (since it is a limit) which gives
$$\frac{d}{dx} f(x)<\frac{d}{dx}g(x).$$ Which establishes your desired property.
Or did i missed out something?
No, take $g(x) = e^{-2x}$ and $f(x) = 2e^{-x}$.
The reason for this is that for $\frac{f(x)}{g(x)}$ to decrease, we must have
$$\frac{f'g - fg'}{g^2} < 0$$
which is if and only if $\frac{f'}{f} < \frac{g'}{g}$ because $f, g$ are positive. Clearly $f' < g'$ is not sufficient since the weights $\frac{1}{f}, \frac{1}{g}$ can change the inequality sign.
In other words, the condition equivalent to the decreasing of $\frac{f}{g}$ is that the relative decrease of $f$ must be faster than the relative decrease of $g$ and the assumption only ensures the above for the absolute decrease of $f$ and $g$.