Suppose that $f(x_1,y_1)f(x_2,y_2) \le f(x_1,y_2)f(x_2,y_1)$ holds for $x_1 \le a \le x_2$ and $y_1 \le b \le y_2$ . Let $(X,Y)$ have the joint density function $f$ and joint distribution function $F$.Suppose that $f(x_1,y_1)f(x_2,y_2) \le f(x_1,y_2)f(x_2,y_1)$ holds for $x_1 \le a \le x_2$ and $y_1 \le b \le y_2$ .Show that $F(a,b) \le F_{x}(a)F_{y}(b) $ where 
$F_{x}$ and $F_{y}$ are marginal dfs.
I tried by finding that RHS is $F_{X}(a)F_{Y}(b)=(F_{X}(x_1)+ \int_{x_1}^{a}f(x,y) dy)(F_{Y}(y_1)+ \int_{y_1}^{b}f(x,y) dx)$
But I don't know how to proceed next
 A: I liked your question, so $+1$.
Let's consider the inequality $f(x_1,y_1)f(x_2,y_2) \leq f(x_2,y_1)f(x_1,y_2)$.
Integrate both sides over corresponding intervals, as follows(involving four variables, so we have to integrate four times). It's all rigorously done, since Fubini's theorem holds.
$$
\int_{-\infty}^a\int_{-\infty}^b\int_{a}^{\infty}\int_{b}^\infty f(x_1,y_1) f(x_2,y_2) \leq  \int_{-\infty}^a\int_{-\infty}^b \int_{a}^{\infty} \int_{b}^\infty f(x_2,y_1) f(x_1,y_2)
$$
Your task : rearrange the integrals on the left and right hand sides so that you get the following inequality:
$$
P(x \leq a, y \leq b) P (x \geq a, y \geq b) \leq P(x \leq a, y \geq b)P(x \geq a, y \leq b)
$$
Once this is done, merely for our convenience, we define:
$$
A := P(x \leq a, y \leq b) ; B := P (x \leq a, y \geq b); C:=P(x \geq a, y \leq b)
$$
Once we have done the above, it's clear that the above statement says $A(1-A-B-C) \leq BC$.
Rearrange all the terms, to get $A \leq A^2 + AB+AC+BC = (A+B)(A+C)$.
The left hand side is $A = F(a,b)$, and the right hand side, if seen carefully, is:
$$
A+B = P(x \leq a) = F_x(a) ; A+C = P(y \leq b) = F_y(b)
$$
Hence, the result follows.
Of course, if you do not get your "task", then do get back, I'm happy to help.
