Simple proof of $\int_I \exp(-f(x))dx\geq \int_I \exp(-g(x))dx$ if $\int_I (f(x) - g(x))\exp(-g(x))dx = 0$ If for two functions $f(x)$ and $g(x)$ we have that:
$$\int_I (f(x) - g(x))\exp(-g(x))dx = 0 \tag{1}$$
where $I$ is an arbitrary interval on the real line, then
$$\int_I \exp(-f(x))dx\geq \int_I \exp(-g(x))dx \tag{2}$$
This follows in a straightforward way from the Bogoliubov inequality. It seems to be a reasonable powerful tool to get to sharp bounds for certain integrals or summations, but it doesn't seem to be used a lot for this purpose in mathematics. I was wondering if a more straightforward proof can be given for the special case of integrals and summations.
This can be used to obtain sharp lower bounds for integrals of the form $\int_I \exp(-f(x))dx$ by writing down a function $g(x)$ containing parameters, and then imposing the constraint (1) to eliminate one parameter and then maximizing $\int_I \exp(-g(x))dx$ w.r.t. the remaining parameters. A simple example is to take $f(x) = x^2$ and $g(x) = a + b x$ for $b>0$ and $I$ the positive real line, which yields the inequality $\pi \geq e$.

Using $e^u \ge 1+u$ for $u \in \Bbb R$ we have
$$
 e^{-f(x)} - e^{-g(x)} = \bigl (e^{g(x)-f(x)} - 1 \bigr) e^{-g(x)} \ge \bigl(g(x)-f(x)\bigr) e^{-g(x)}
$$
so that
$$
 \int_I e^{-f(x)} \, dx - \int_I e^{-g(x)} \, dx \ge \int_I \bigl(g(x)-f(x)\bigr) e^{-g(x)} \, dx \, .
$$
If $(1)$ holds then the right-hand side is zero, and that implies $(2)$.
We can even conclude that strict inequality holds in $(2)$, unless $f(x) = g(x)$ for almost all $x \in I$.
