Let $f:\mathbb{R} \to \mathbb{R}$ and $g:\mathbb{R} \to \mathbb{R}$ be two differentiable and strictly increasing functions. Consider two statements:
- $\frac{f(x)}{g(x)}$ is strictly increasing in $x$.
- $f(x)-g(x)$ is strictly increasing in $x$.
My question is what is the relationship between (1) and (2)? Are they equivalent or does one imply another?
Since $f$, $g$ are strictly increasing, $(2)$ implies that $$f'(x)>g'(x)>0 \implies 1>\frac{g'(x)}{f'(x)}$$ while $(1)$ is equivalent to $$f'(x)>\frac{f(x)}{g(x)}g'(x) \implies g(x)>f(x)\frac{g'(x)}{f'(x)}$$ For $x$ such that $f(x)<0$ (if such $x$ exists) you can rewrite the last expression as $$\frac{g(x)}{f(x)}<\frac{g'(x)}{f'(x)}<1$$ So, can you find $f,g$ that violate this, while satisfying all the previous?
For example $f(x)=2x+1$ and $g(x)=x$ are both strictly increasing in $\mathbb R$, their difference is clearly $f(x)-g(x)=x+1$ strictly increasing but $$\frac{f(x)}{g(x)}=2+\frac1x$$ is decreasing in its domain.
Edit: The direction $(1) \implies (2)$ is not true either. Consider for example $f(x)=2x$ and $g(x)=3x+1$. Both are strictly increasing, $f(x)/g(x)$ is strictly increasing in its domain, but $f(x)-g(x)=-x-1$ is strictly decreasing.