Product of two independent, random draws is usually less than the square of a random draw. Why? Suppose a uniform distribution, D over positive real numbers. On average, the product of two independent draws from D is less than the square of a random draw from D. I wrote a MWE in R code that simulates the above. My question is, why?

n_sim = 1000000
min_val = 1
max_val = 10000000000
a = runif(n   = n_sim,
          min = min_val, max = max_val)
b = runif(n   = n_sim,
          min = min_val, max = max_val)
c = runif(n   = n_sim,
          min = min_val, max = max_val)
table(a&lt;b)
table(a*b &lt; c^2)
Your finding is true for any distribution, positive or not, uniform or not. Let $X$ and $Y$ be independent, each with the same distribution as $Z$. Your observation is that
$$E(XY -Z^2)\le0.\tag1$$
Since $X$ and $Y$ are independent, the LHS of (1) equals $E(X)E(Y)=[E(Z)]^2$ so asserting (1) is the same as asserting
$$
[E(Z)]^2-E(Z^2)\le0.\tag2
$$
You can rearrange (2) to the form
$$
E(Z^2)-[E(Z)]^2\ge0,\tag3$$
which is a true statement for any $Z$ because the LHS of (3) equals $E[(Z-E(Z))^2]$, which is the expectation of a non-negative random variable (it's also the variance of $Z$).
Your finding is true for any distribution, positive or not, uniform or not. Let $X$ and $Y$ be independent, each with the same distribution as $Z$. Your observation is that
$$E(XY -Z^2)\le0.\tag1$$
Since $X$ and $Y$ are independent, the LHS of (1) equals $E(X)E(Y)=[E(Z)]^2$ so asserting (1) is the same as asserting
$$
[E(Z)]^2-E(Z^2)\le0.\tag2
$$
You can rearrange (2) to the form
$$
E(Z^2)-[E(Z)]^2\ge0,\tag3$$
which is a true statement for any $Z$ because the LHS of (3) equals $E[(Z-E(Z))^2]$, which is the expectation of a non-negative random variable (it's also the variance of $Z$).