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.



Elementary inequality about integrals of exponentials of concave functions (possibly connected to log concave distributions)
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 $$\int\_{\mathbb R} f(x) e^{-f(x)} dx
=1-b<0.$$


