Prove that $(a+b+c-d)(a+c+d-b)(a+b+d-c)(b+c+d-a)\le(a+b)(a+d)(c+b)(c+d)$ 
Let $a,b,c,d>0$. Prove that $$(a+b+c-d)(a+c+d-b)(a+b+d-c)(b+c+d-a)\le(a+b)(a+d)(c+b)(c+d)$$

I don't know how to begin to solve this problem

Also we can use the following reasoning.
Since $\prod\limits_{cyc}(a+b)-(a+b+c+d)(abc+abd+acd+bcd)=(ac-bd)^2\geq0$,
it remains to prove that
$$(a+b+c+d)(abc+abd+acd+bcd)\geq\prod\limits_{cyc}(a+b+c-d)$$
which is also true by BW.
