<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] Let $x^2+y^2=m$.
Thus, $$(m+6)m\geq A(m-2)$$ or
$$m^2+(6-A)m+2A\geq0,$$
for which we need $$(6-A)^2-8A\leq0,$$
which gives $2\leq A\leq18$.
For $A=18$ we get $m=6$ or $x^2+y^2=6$, which says that the equality occurs.
Thus, $18$ is our answer.
Done!
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