Answer Should be Option (D).
Lets assume $X$ is a random variable uniformly distributed in $[0,1]$ and $Y$ is $1-X$ .
Lets assume $X$ is a random variable uniformly distributed in $[0,1]$ and $Y=X$.
So, $E(X)=\frac{1}{2}$
$E(Y)=\frac{1}{2}$
Now Given $Z=XY$
So,$E(Z)=E(XY)=\int_{0}^{1} x*x \,dx=\frac{1}{3}$
So, $E(Z) \geq E(X)E(Y)$
[Note here Covariance is positive]
Now , why does option (D) is correct ?
=> It is given , $X$ and $Y$ takes value in the range $[0,1]$.
So, we can write , $0 \leq Y \leq 1$ .
Multiply this inequality by $X$.
So, $0 \leq XY \leq X$
Now take expectation of this Inequality,
So, $E(0) \leq E(XY) \leq E(X)$
So, $0 \leq E(XY) \leq E(X)$
Which proof the option (D) .
Note this inequality Holds true because $X$ and $Y$ takes value in the range between $[0,1]$.