Proving Quadratic Inequality: (x-y)^2 ≥ 0

[sg001]

Homework Statement

By expanding (x-y)^2, prove that x^2 +y^2 ≥ 2xy for all real numbers x & y.

Homework Equations

The Attempt at a Solution

expanding (x-y)^2

x^2 - 2xy + y^2= 0

Hence, x^2 + y^2 = 2xy

But where does the ≥ come into it? and why?
when you put values in (except i which is not real of course) they all come out as = 2xy, which does satisfy ≥2xy, but why does this come into it??
Some insight would be fantastic!

 

[Dick]

If x and y are real then (x-y) is real. (x-y)^2>=0, yes?

 

[sg001]

Ohhhh I see...
Thanks!

 

[Vonalak]

Continuing from Dick's hint:

(x-y)^2 ≥ 0 (Trivial inequality)
x^2-2xy+y^2 ≥ 0
And adding 2xy to both sides, we get:
x^2+y^2 ≥ 2xy as desired.

And if you're up for it, try proving the two variable case of the AM-GM inequality from here (it's pretty simple).