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 $a=\frac{x^2}{z},\, b=\frac{y^2}{z}$ 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.
