Using the Arithmetic Mean-Geometric Mean Inequality. Let a, b, c be real numbers. Prove that:
$(a+b-c)^{2}(b+c-a)^{2}(c+a-b)^{2}\geq(a^{2}+b^{2}-c^{2})(b^{2}+c^{2}-a^{2})(c^{2}+a^{2}-b^{2}) $

We can assume that $\prod\limits_{cyc}(a^2+b^2-c^2)\geq0$.
If $a^2+b^2-c^2<0$ and $a^2+c^2-b^2<0$ so $2a^2<0$, which is contradiction.
Thus, we can assume $a^2+b^2-c^2\geq0$, $a^2+c^2-b^2\geq0$ and $b^2+c^2-a^2\geq0$.
Since $(a+b-c)^2(a+c-b)^2-(a^2+b^2-c^2)(a^2+c^2-b^2)=2(b-c)^2(b^2+c^2-a^2)\geq0$, 
we obtain:
$$\prod\limits_{cyc}(a+b-c)^2(a+c-b)^2\geq\prod\limits_{cyc}(a^2+b^2-c^2)(a^2+c^2-b^2),$$
which is
$$\prod\limits_{cyc}(a+b-c)^2\geq\prod\limits_{cyc}(a^2+b^2-c^2)$$
Done!
