<s>[INST] 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
 [/INST] By way of simplifying the situation, let x = ab and y = a+b. Note that as a and b range over all positive reals, x and y range over all pairs for which x>0 and y > 2 √x because if x lies on the hyperbola ab = x in the first quadrant of the a-b plane, then the least value for a+b will be when a=b and then it is  2 √x. So as long as y > 2 √x there will be positive a and b satisfying the equations between x,y and a,b.
Expressing the given  inequality in terms of x and y we get:
x^2 (y^2 – 2x – 2) > y (x-1), which can be transformed into 
y^2 + y(1-x)/x^2  > 2x+2.
Let u = (1-x)/2x^2  and v = 2x+2, so we require that
y^2 + 2uy > v, which holds iff y > √(u^2 + v) – u (since y>0).
The only way for this condition to hold for all x, y as described above is for
2 √x >= √(u^2 + v) – u.
Move the u to the other side and square to get
u √x >= v/4 – x.
Now replace u and v with their definitions to get the condition:
(1-x)/2x^2  * √x >= (1+x)/2 – x = (1-x)/2.
If 1-x > 0 this simplifies to x<=1 which is true and if 1-x <= 0 it simplifies to x>=1, also true, which proves the inequality.
To clarify things a little, one starts with an arbitrary positive a and b, then finds x, y, u, v. Since y > 2 √x which itself is >= √(u^2 + v) – u, also y > √(u^2 + v) – u. This last  inequality then can be traced back to demonstrating the original inequality.
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