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  

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.
