Question

Prove that if x and y are distinct and positive real numbers then (x/y) + (y/x) > 2. discuss the conditions (on x and y) under which this inequality becomes an equality.

Answer 1

1.

$\dfrac{x}{y}+\dfrac{y}{x} > 2|\cdot xy\\ x^2+y^2>2xy\\ x^2-2xy+y^2>0\\ (x-y)^2>0$

$(x-y)^2>0$ is true for any two real numbers such that $x\not =y$, all the more for positive ones.

2.

$x=y$