Integral inner product with exponential function

Question

Suppose on some unknown interval $[0, I]$ we have non-negative functions $f, g : [0, I] \rightarrow \mathbb{R}^{\geq 0}$. If we know that \begin{aligned} \int_0^I f & = c \\ \int_0^I e^f & = e^c \\ \int_0^I g e^f & = d e^c, \end{aligned}

is it possible to deduce that $$\int_0^I gf \leq d c \ ?$$

I expect that $I=1$ and $f=c$ constant is the case where it is tight, where it holds trivially. I have so far that $I (c - \ln{I}) \geq c$ from applying Jensen's inequality to the first two lines, which gives $1 \leq I < c$.

If $I=1$ then Jensen's tells that the $f$ should be constant, so we're done. I have made calculations that suggested this holds for $I>1$ too, but they are far from a proof.

Is the conclusion even true? I've never done functional inequalities like this before.

Answer 1

Fix $f$, then your claim is just $\int_0^I g (e^{f-c}-f/c)dx\geqslant 0$. This should hold for every positive $g$, so, the function $h:=e^{f-c}-f/c$ must be non-negative. Since $\int_0^Ih=0$, this simply means that $h\equiv 0$. But this is not always the case.