Ratio of convex functions with dominating derivatives is convex?

I think the answer is no necessarily. Note that

$$ f=g+ h mbox{ where } h^{(n)}(x)ge0 ;; forall n ge 0 $$ And $frac{f}{g}$ is convex iff $frac{h}{g}$ is convex.

Now suppose $g(x)=e^x$. Then we need $h$ such that $h^{(n)}(x)ge0 ; forall n$ and

$$ left(frac{h}{e^x}right)''=frac{h''(x)-2h'(x)+h(x)}{e^x}<0 mbox{ somewhere } $$ Set $h(x)=ax+b$, where $a$ and $b$ are positive constants. Then $$ h''(x)-2h'(x)+h(x)= 0-2a+ax+b=a(x-2)+b $$ which is negative in $x=2-frac{b}{a}-varepsilon$.

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I think that given any $g$, we can choose $h$ such that

$$ left(frac{h}{g}right)''<0 $$ but surely it's a little more difficult to prove.