prove that : if $a, b \in \mathbb{R}^+$ : then : $a^2b^2(a^2+b^2-2)\geq (a^2+b^2)(ab-1)$ prove that : if $a, b \in \mathbb{R}^+$ : then :
$$a^2b^2(a^2+b^2-2)\geq (a^2+b^2)(ab-1)$$

$$a^4b^2+b^4a^2-2a^2b^2 \geq a^3b-a^2b+b^2a-a^2$$
$$a^4b^2+b^4a^2-2a^2b^2 \geq a^3b-a^2+b^3a-b^2$$
now what ?

Let $a+b=2u$ and $ab=v^2$.
Thus, we need to prove that
$$v^4(4u^2-2v^2-2)\geq(4u^2-2v^2)(v^2-1)$$ or
$$v^4(2u^2-v^2-1)\geq(2u^2-v^2)(v^2-1)$$ or
$$2u^2v^4-v^6-v^4\geq2u^2(v^2-1)-v^4+v^2$$ or
$$2u^2(v^4-v^2+1)\geq v^6+v^2.$$
But $v^4-v^2+1>0$ and $u^2\geq v^2$ because it's just $(a-b)^2\geq0$.
Thus, it's enough to prove that
$$2v^2(v^4-v^2+1)\geq v^6+v^2,$$
which is
$$v^2(v^2-1)^2\geq0.$$
Done!
