prove that $a^2 b^2 (a^2 + b^2 - 2) \ge (a + b)(ab - 1)$ Good morning
help me to show the following inequality
for all $a$, $b$ two positive real numbers
$$a^2 b^2 (a^2  + b^2  - 2) \ge (a + b)(ab - 1)$$
thanks you  
 A: Let $a+b=2u$ and $ab=v^2$, where $v>0$. Hence, we need to prove that $2v^4u^2-(v^2-1)u-v^6-v^4\geq0$, for which it's enough to prove that $u\geq\frac{v^2-1+\sqrt{(v^2-1)^2+8v^4(v^4+v^6)}}{4v^4}$ or $(4v^5-v^2+1)^2\geq(v^2-1)^2+8v^4(v^4+v^6)$ because $u\geq v$, or $(v-1)^2(v+1)(v^2+v+1)\geq0$. Done!
A: I found an AM-GM proof!
By AM-GM$,$ we have$:$
$${a}^{2}b\leq \frac25\,{a}^{4}{b}^{2}+\frac25\,a+\frac15\,b,$$
$$a{b}^{2}\leq \frac{1}{10}\,{a}^{4}{b}^{2}+\frac{3}{10}\,{a}^{2}{b}^{4}+\frac{3}{5}\,b,$$
$$2\,{a}^{2}{b}^{2}\leq \frac12{a}^{4}{b}^{2}+{\dfrac {7}{10}}{a}^{2}{b}^
{4}+\frac35a+\frac15b.$$
Summing them we get the result.
Done.
A: Let $a=\frac{x^2}{z},\, b=\frac{y^2}{z}$ then $x,y \geqq 0$ and $z>0.$ Then we have$:$
$$z^6 \cdot  (\text{LHS}-\text{RHS})={x}^{3}{y}^{3} \left( x-y \right) ^{2} \left( {x}^{3}y+{x}^{2}{y}^{2}+ x{y}^{3}+{z}^{2} \right) + \left( xy+z \right)  \left( {x}^{2}{y}^{2}+ xyz+{z}^{2} \right)  \left( xy-z \right) ^{2} \left( {x}^{2}+{y}^{2} \right) \geqq 0$$
Another proof$,$ you can see here.
