The following problem is from Rohatgi Let (X,Y) have 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 any $x_1 \le a \le x_2$ and $y_1 \le b \le y_2 $. Show that $F(a,b) \le F_1(a)F_2(b)$.
My attempt:
I tried writing it out from definition: $$F(a,b) = P(X \le a, Y \le b) = \int_{-\infty}^a \int_{-\infty}^bf(x,y)dydx$$ but I'm not sure how to introduce the given inequality
Any hints?
Hint: \begin{align} \int_{-\infty}^a\int_b^\infty f(x,y)dydx&=\int_{-\infty}^a\left[\int_{-\infty}^\infty f(x,y)dy-\int_{-\infty}^bf(x,y)dy\right]dx\\ &=F_1(a)-F(a,b)\ ,\\ \int_a^\infty\int_{-\infty}^bf(x,y)dydx&=F_2(b)-F(a,b)\ \text{, and}\\ \int_a^\infty\int_{b}^\infty f\left(x,y\right)dydx&=1-F_1(a)-F_2(b)+F(a,b)\ , \end{align} and integrating the inequality you're given over the ranges $\ -\infty < x_1\le a\ $, $\ -\infty < y_1\le b\ $, $\ a < x_2\le \infty\ $, and $\ b < y_2\le \infty\ $ gives $$ \int_{-\infty}^a\int_{-\infty}^b f\left(x_1,y_1\right)dy_1dx_1\int_a^\infty\int_{b}^\infty f\left(x_2,y_2\right)dy_2dx_2\le\\ \int_{-\infty}^a\int_b^\infty f\left(x_1,y_2\right)dy_2dx_1\int_a^\infty\int_{-\infty}^bf\left(x_2,y_1\right)dy_1dx_2\ . $$ Can you take it from there?