Elementary inequality about integrals of exponentials of concave functions (possibly connected to log concave distributions)

Question

I think the following inequality might be true and was hoping somebody might spot it or know a proof:

Suppose $f:\mathbb R\to \mathbb R$ is convex and suitably nice so that $$\int_{\mathbb R} e^{-f(x)} dx = 1$$ Then is it true that $$\int_{\mathbb R} f(x) e^{-f(x)} dx\ge 0$$ ?

I also wonder what area of mathematics this might fit into or be a baby case of (perhaps the theory of logarithmically concave distributions)?

Thanks in advance.

Answer 1

The answer is no.

E.g., take any real $b>1$ and let $a:=2e^b$. Let $f(x):=a|x|-b$ for all real $x$.

Then $\int_{\mathbb R}e^{-f(x)} dx=1$ but $$\int_{\mathbb R} f(x) e^{-f(x)} dx =1-b<0.$$