Geometric Inequality $(a+b)(b+c)(c+a)(s-a)(s-b)(s-c)\leq (abc)^{2}$ everyone.   
$a$,$b$,$c$ are three sides of a triangle.
Prove or disprove the following.   
$(a+b)(b+c)(c+a)(a+b-c)(b+c-a)(c+a-b)\leq 8(abc)^{2}$   
I know two inequalities.
$8(s-a)(s-b)(s-c)\leq abc~$ , $~(a+b)(b+c)(c+a)\geq 8abc$   
But for the above combination of them, I have no idea.
Thanks in advance.

Use Heron's formula
$$(a+b-c)(b+c-a)(c+a-b)(a+b+c)=\dfrac{a^2b^2c^2}{R^2}$$
where $R$ be the center of the circumcircle of $\Delta ABC$.
your inequality can write as
$$8R^2\ge\dfrac{(a+b)(b+c)(a+c)}{a+b+c}$$
since
$$9R^2\ge a^2+b^2+c^2$$
it suffices to prove
$$8(a+b+c)(a^2+b^2+c^2)\ge 9(a+b)(b+c)(a+c)\tag{1}$$
use AM-GM  inequality
$$27(a+b)(b+c)(a+c)\le 8(a+b+c)^3$$
then it easy to prove (1)
