Question

prove For all real numbers x and y, if x ≠ y, x > 0, and y > 0, then x/y+ y/x>2

Answer 1

Here's what I have so far:

Since x and y are positive real numbers, xy is positive and we can multiply both sides of the inequality by xy to obtain
(x/y+ y/x)∙xy >2 ∙xy
x^2+ y^2 >2xy
By combining all terms on the left side of the inequality, we see that x2 - 2xy + y2 > 0, and then by factoring the left side, we obtain (x - y)2 > 0

. Solving this inequality we get x > y. Since x > 0, and y > 0, then x/y>0 and y/x>0.