Geometric Inequality $(a+b)(b+c)(c+a)(s-a)(s-b)(s-c)\leq (abc)^{2}$

Question

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.

Answer 1

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)