Find the largest possible real number $C$ such that for all pairs $(x, y)$ of real numbers with $x \ne y$ and $xy = 2$, $$\frac{((x + y)^2- 6))(x-y)^2 + 8))}{(x-y)^2} \ge C.$$Also determine for which pairs $(x, y)$ equality holds.
Problem
Source: New Zealand NZMOC Camp Selection Problems 2013 p10
Tags: algebra, inequalities
sqing
03.10.2021 04:52
Let $x,y$ be real numbers such that $x \ne y$ and $xy = 2$. Prove that $$\frac{((x + y)^2- 6)((x-y)^2 + 8)}{(x-y)^2} \ge18$$Equality holds when $(x,y)=(-1-\sqrt 3, 1-\sqrt 3)$ or $(x,y)=(\sqrt 3-1, \sqrt 3+1).$
sqing
03.10.2021 05:31
Let $x,y$ be real numbers such that $x \ne y$ and $xy = 2$. Prove that $$\frac{((x + y)^2- 4)((x-y)^2 +4)}{(x-y)^2} \ge20$$Equality holds when $(x,y)=(-1-\sqrt 3, 1-\sqrt 3)$ or $(x,y)=(\sqrt 3-1, \sqrt 3+1).$ $$ \frac{((x + y)^2- 7)((x-y)^2 +1)}{(x-y)^2}\ge 4$$Equality holds when $(x,y)=(-2, -1)$ or $(x,y)=(1, 2).$ $$ \frac{((x + y)^2- 1))((x-y)^2 +7))}{(x-y)^2} \ge28$$
sqing
06.10.2021 11:17
sqing wrote: Let $x,y$ be real numbers such that $x \ne y$ and $xy = 2$. Prove that $$\frac{((x + y)^2- 6)((x-y)^2 + 8)}{(x-y)^2} \ge18$$Equality holds when $(x,y)=(-1-\sqrt 3, 1-\sqrt 3)$ or $(x,y)=(\sqrt 3-1, \sqrt 3+1).$
Attachments:
