prove (x/y) + (y/x) > or = 2?

Question

if x>0 and y >0 prove that:

(x/y) + (y/x) > or = 2

Answer 1

Multiply the inequality by xy:

x^2 + y^2 ≥ 2xy

x^2 + y^2 - 2xy ≥ 0

(x - y)^2 ≥ 0

This last is always true since any real quantity squared is ≥ 0

Answer 2

(x/y) +(y/x) ≥ 2
the number for x or y cannot be zero because dividing any number is meaning less.
The minimum integer that can be taken is
x = 1 and y =2
so (1/2) + (2/1)
=.5 +2
= 2.5

OR

x = 2 and y =1
so
(2/1) + (1/2)
=2 +.5
= 2.5
so it ≥ 2