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

The simplest method to solve this question should be the following:
Let 
$$b+c+d-a=w\tag{1}$$
$$a+c+d-b=x\tag{2}$$
$$a+b+d-c=y\tag{3}$$
$$a+b+c-d=z\tag{4}$$
So, the given inequality is equivalent to 
$$16wxyz \leq (w+x)(x+y)(y+z)(z+w)\tag{*}$$

It is not hard to see that at most one of $w,x,y,z$ is less than zero.
So, there are two cases to consider:
$(1)$ All of $w,x,y,z \geq 0$. 
The result is true by AM-GM Inequality
Equality holds when $a=b=c=d$
$(2)$ One of $w,x,y,z$ , say $w<0$ and $x,y,z \geq 0$
$\text{L.H.S } < 0$, while $\text{R.H.S } > 0$
So, inequality holds.
