prove that $ a^2 b^2 \left( {a^2 + b^2 - 2} \right) \ge \left( {a + b} \right)\left( {ab - 1} \right) $ good evening I want to show that
$(1)a;b\in\mathbb {R^*_+}:a^2 b^2 \left( {a^2  + b^2  - 2} \right) \ge \left( {a + b} \right)\left( {ab - 1} \right)
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$
\begin{array}{l}
 \frac{a}{b} + \frac{b}{a} \ge 2 \\ 
 \frac{a}{b} + \frac{b}{a} = \frac{{a^2  + b^2 }}{{ab}} \Rightarrow a^2  + b^2  = ab\left( {\frac{a}{b} + \frac{b}{a}} \right) \ge 2ab \\ 
 a^2 b^2 \left( {a^2  + b^2  - 2} \right) \ge a^2 b^2 \left( {2ab - 2} \right) = 2a^2 b^2 \left( {ab - 1} \right) \\ 
 \end{array}
$
$ x;y;z\in\mathbb {R^*_+}:$
$(2)\sum_{cyc}^{ } xy(x+y-z)\ge \sqrt {3(x^3y^3+y^3z^3+z^3x^3)}$
thank you in advance

Do you keep the first true in the case of 
$a;b\in\mathbb {R^*_+} and :   a^2  + b^2  \ge 2$
$ (2)x;y;z\in\mathbb {R^*_+}:$
$\sum_{cyc}^{ } xy(x+y-z)\ge \sqrt {3(x^3y^3+y^3z^3+z^3x^3)}$
Thank you for all helping and guidance. ...
