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

Question

prove that : if $a, b \in \mathbb{R}^+$ : then :

$$a^2b^2(a^2+b^2-2)\geq (a^2+b^2)(ab-1)$$

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$$a^4b^2+b^4a^2-2a^2b^2 \geq a^3b-a^2b+b^2a-a^2$$

$$a^4b^2+b^4a^2-2a^2b^2 \geq a^3b-a^2+b^3a-b^2$$

now what ?

Answer 1

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 $a=b=1$.

Answer 2

it is equivalent to $$(a-b)^2((ab)^2-ab+1)+2ab(ab-1)^2\geq 0$$ which is true.

Answer 3

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!