<s>[INST] 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 ?
 [/INST] Let $ab=x$, $a^2+b^2=y$, then $0\leq 2x\leq y$ and we want
$$x^2(y-2)\geq y(x-1)$$
$$x^2y-2x^2\geq xy-y$$
$$y\geq\frac{2x^2}{x^2-x+1}$$
It is equivalent to show
$$2x\geq\frac{2x^2}{x^2-x+1}$$
$$x^2-x+1\geq x$$
$$(x-1)^2\geq0$$
Equality holds iff $a=b=1$.
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