Inequality for the division of expectations of convex functions

Question

This strongly relates to my other Question, but I am gonna formulate it in a simpler and more general way.
Suppose $X$ is a real random variable, $\phi(x)$ is a convex function and the functions $g$, $f$ are such that $g(x) \le f(x)\ \forall x\in\mathbb{R}$. Does the following inequality hold?\begin{equation}
\frac{\phi\big(E[f(X)]\big)}{\phi\big(E[g(X)]\big)}\le \frac{E\big[\phi\big(f(X)\big)\big]}{E\big[\phi\big(g(X)\big)\big]}
\end{equation}
The problem here is that, because of Jensen's Inequality, both the enumerator and the denominator grow when $\phi$ is inside the expectations. So the change in magnitude of the enumerator has to be larger than the demoninator's.
So for example let $X\in\{1,2\}$ and $p=0.5$ for both events. Let $\phi(x)=e^x,\ g(x)=x,\ f(x)=x+1$. In this case both sides of the equation work out to be the same.
I tried out some other examples with simple two or three-valued r.v. and the inequality always held.
My intuition for this is, since $\phi$ is convex, the larger function causes a larger change in magnitude.

Answer 1

Not true. For example if $f(x)=1$ for all $x$ and $\phi $ is strictly convex with $\phi (1)=1$ ( say $\phi (x)=e^{x}-e+1$) then the inequality goes in the opposite direction and you can easily give examples where the reverse inequality is strict.