<s>[INST] An olympic mathematics problem regarding Cauchy-Schwarz This question was asked in Turkish National Maths Olympiad in 2008.

For all $xy=1$ we have $((x+y)^2+4)\cdot ((x+y)^2-2) \ge A\cdot(x-y)^2$.
What is the maximum value $A$ can get?

My efforts regarding this problem;
$(x+y)^2-8 \ge A\cdot(x-y)^2$
Using the property $xy=1$ ;
$x^2+y^2-6\ge A\cdot (x^2+y^2-2)$
$\sqrt{\dfrac{x^2+y^2}{2}} \ge\sqrt{\dfrac{A\cdot(x^2+y^2-2)}{2}}$
Therefore $\sqrt{\dfrac{A\cdot(x^2+y^2-2)}{2}}=\dfrac{x+y}{2}$
Although moving further that doesn't work I applied Cauchy-Schwarz by the way if I didn't mention it before.
How should I proceed?
What are you suggestions?
 [/INST] Note that your first expression:

$$(x+y)^2-8\ge A\cdot (x-y)^2$$

must be:
$$(x+y)^4+2(x+y)^2-8\ge A\cdot (x-y)^2.$$
Alternative solution: You can also denote:
$$(x+y)^2=z; \ \ \ (x-y)^2=(x+y)^2-4xy=z-4.$$
Then:
$$(z+4)(z-2)\ge A\cdot (z-4) \Rightarrow$$
$$z^2+(2-A)z+4A-8\ge 0$$
This inequality is true for all $z$ (all $x,y$) if:
$$(2-A)^2-4(4A-8)\le 0 \Rightarrow (A-2)(A-18)\le 0 \Rightarrow 2\le A\le 18.$$
Hence the maximum value of $A$ is $18$.
P.S. Was there any other condition, like $x\ne y$? Otherwise if $x=y=1$, the left hand side is $16$, while the right hand side is $0$, implying $A$ can be any number, hence no max.
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