Inequality involving taking expectations

There are three convexly decreasing functions $f, g, h:\mathbb{R^+}\rightarrow \mathbb{R^+}$, and $f(x)h(x)<1$ for $\forall x$.
I need to prove that

$E[f(x)^2]E[g(x)h(x)]<E[f(x)g(x)]\left(1+E\left[f(x)h(x)\right]\right)$

for an arbitrary probability distribution.

Solutions Collecting From Web of "Inequality involving taking expectations"

The result is true more generally without convexity.

Fact 1: If $r(X)$ and $q(X)$ are non-decreasing functions of a random variable $X$, then their covariance is non-negative. See link here for this fact:
covariance of increasing functions

Fact 2: If $m(X)$ and $n(X)$ are non-increasing functions of a random variable $X$, then their covariance is also non-negative, since:

$$ E[m(X)n(X)] = E[(-m(X))(-n(X))] \geq E[-m(X)]E[-n(X)] = E[m(X)]E[n(X)] $$

where the inequality follows from Fact 1 together with the observation that $-m(X)$ and $-n(X)$ are non-decreasing.

We can apply Fact 2 to your problem: The functions $f(X)^2$ and $g(X)h(X)$ are both non-increasing in $X$, and so by Fact 2:

$$ E[f(X)^2]E[g(X)h(X)] \leq E[f(X)^2g(X)h(X)] \leq E[f(X)g(X)] $$

where the final inequality uses the fact that $f(X)h(X)\leq 1$ for all $X$. The final term is less than it would be if we add the positive value $E[f(X)g(X)]E[f(X)h(X)]$.


Your previous question shows the result is not true if we remove the condition $f(X)h(X)\leq 1$. That link is:

inequality involving taking expectation