# An olympic mathematics problem regarding Cauchy-Schwarz

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?

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!

• Wait, why $(6-A)^2 - 8A \leq 0$ and not $\geq$? A negative discriminate seems less than ideal. Or is that derived from something different?
– Yakk
Aug 30, 2017 at 19:18
• Rather, for $m^2 + (6-A)m + 2A$ to be $\geq 0$ for all $m$, we must have exactly one root, and it must be convex up (as it obviously is). As such we should simply directly solve for $(6-A)^2 - 2A = 0$ with no inequality at all.
– Yakk
Aug 30, 2017 at 19:25
• @Yakk I think it's better to write $(6-A)^2-8A\leq0$ because for $(6-A)^2-8A<0$ the inequality $m^2+(6-A)m+2A\geq0$ is also true for all reals $m$. Aug 30, 2017 at 20:01
• Ah yes, having no roots is also a valid choice. I missed that.
– Yakk
Aug 30, 2017 at 20:24

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