Prove or find a counterexample $\forall x>0: f(2x)-f(x)<g(3x)-g(2x)$, given information about $f, g$.

Question

Let $f,g:(0, \infty) \rightarrow \mathbb{R}$ be two functions that satisfy the following for all $x>0$: $g'(x)>f'(x)$ and $f''(x)>0$.

Prove or find a counterexample: $$ \forall x>0: f(2x)-f(x)<g(3x)-g(2x). $$

I've tried this for a long time, but I didn't make much progress.
I tried to move everything to one side and then take derivative.

If we move everything to the LHS and take derivative, we get $2f'(2x)-f'(x)+2g'(2x)-3g'(3x)$.

I don't know if this is positive or negative, and even if I knew, I don't see how it helps me. (It would help to know it if $f,g$ were defined at $0$).

Also I did not find any counterexample. Can someone please help me? And please write in your answer what was your intuition for this problem and why I still have no intuition to determine if the claim is true or false.

Thanks.

Answer 1

There exists $c\in (x,2x)$ such that $$f(2x)-f(x)=f'(c)x.$$

There exists $d\in (2x,3x)$ such that $$g(3x)-g(2x)=g'(d)x.$$

So, we need to show that $f'(c)<g'(d).$

Now, $f''(x)>0$ implies that $f'(x)$ is increasing. So we have, using that $c<d$ and $f'(x)<g'(x),$ that

$$f'(c)<f'(d)<g'(d)$$ and we are done.