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

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Homework Help Overview

The problem involves proving the quadratic inequality (x-y)² ≥ 0, which leads to the conclusion that x² + y² ≥ 2xy for all real numbers x and y. The discussion centers around understanding the implications of this inequality and its derivation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to expand (x-y)² and questions how the inequality arises from the equality x² + y² = 2xy. Other participants affirm the non-negativity of (x-y)² and discuss the implications of this trivial inequality.

Discussion Status

Participants are engaging with the problem, with some providing insights into the derivation of the inequality. There is an acknowledgment of the trivial nature of the inequality (x-y)² ≥ 0, and a suggestion to explore further implications, such as the AM-GM inequality.

Contextual Notes

There is a focus on the conditions under which the inequality holds, specifically for real numbers, and some participants express confusion about the transition from equality to inequality.

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