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  

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!
