Quadratic inequality, proof.

  • Thread starter sg001
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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!
 

Answers and Replies

  • #2
Dick
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If x and y are real then (x-y) is real. (x-y)^2>=0, yes?
 
  • #3
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Ohhhh I see...
Thanks!
 
  • #4
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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).
 

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