This is equivalent to ask $cov(X, 1/X)$ when $X \in[2, 3]$.
Can you do this with some clever method?
Edit: clearly $E(X)E(Y) = E(XY) - cov(X,Y) = 1 - cov(X,Y) = 1- cov(X, 1/X)$,
clearly, $X$ and $\frac{1}{X}$ will move in opposite direction, so $cov(X, 1/X) \le 0$,
thus the minimum of $E(X)E(Y)$ will be 1. But how to get the maximum?
My guess is that for example, $X$ could be 50% equals to 2, and 50% equals to 3, $E(X) = 2.5$.
thus $1/X$ will half time 1/2, half time 1/3. $E(1/X) = 5/12$;
Thus it's easy to get $E(X)E(Y) = 25/24$. But is this the maximum? Seems to be by instinct. But how do you proove it?
The maximum is indeed $25/24$.
We want to maximize $E[X]E[1/X]$ over all random variables $X$ that take values in the interval $[2,3]$.
Fact 1: We have
$$ (1/x) + (1/6)x \leq 5/6 \quad \forall x \in [2,3]$$ with equality when $x \in \{2,3\}$.
Fact 2: Fix $m \in [2,3]$. To maximize $E[1/X]$ over all random variables $X$ that take values in $[2,3]$ and that satisfy $E[X]=m$, we take the random variable $X^*$ that satisfies $$P[X^*=2]=p, P[X^*=3]=1-p$$ where $p$ is the value in $[0,1]$ that satisfies $$m=2p+3(1-p)$$ Proof: Let $X$ be a random variable that takes values in $[2,3]$ and has mean $m$. Then we have by Fact 1: $$ (1/X) + (1/6) X \leq 5/6 = (1/X^*) + (1/6)X^*$$ Taking expectations of the above gives: $$ E[1/X] + (1/6) E[X] \leq E[1/X^*] + (1/6)E[X^*]$$ Since we know $E[X]=E[X^*]$ we obtain: $$ E[1/X]\leq E[1/X^*] \quad \Box$$
By Fact 2, it suffices to maximize $E[X]E[1/X]$ over all random variables of the type $X^*$ that satisfy $P[X^*=2]=p, P[X^*=3]=1-p$. So $$E[X^*]E[1/X^*] = (2p + 3(1-p))((1/2)p + (1/3)(1-p))$$ This is a quadratic equation in $p$ that is maximized at $p^*=1/2$, from which we obtain $$ E[X^*]E[1/X^*]=25/24$$