How to Approach Deriving Inequalities?

AI Thread Summary
To derive the inequality x² + xy + y² ≥ 0, consider the equation x² + xy + y² = 0 and analyze its real solutions. If real solutions exist, they define a curve that separates the plane into regions where the inequality holds true. If no real solutions exist, then the inequality must be satisfied for all points in the plane. The discussion emphasizes the importance of understanding the behavior of the equation to approach the derivation of the inequality effectively. This method highlights the relationship between the equation and the regions defined by the inequality.
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Homework Statement



Derive the following inequality.

Homework Equations



x^{2}+xy+y^{2} \geq 0

The Attempt at a Solution



I don't know how to get started. How do you derive inequalities?

I'm not looking for the answer, just general tips.
 
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That term on the left looks similar to (x+y)(x+y). How could you use that?
 
(x+y) squared has +2xy as its middle term
 
I don't think the similarity (or not) to (x + y)(x + y) is any help.

Think about the equation x2 + xy + y2 = 0.
Are there any real solutions to this equation?

If yes, then the real solutions (x, y) are the graph of a curve that separates the portion of the plane for which x2 + xy + y2 > 0 from the other portion of the plane where x2 + xy + y2 < 0.

If no, then all points (x, y) in the plane must satisfy exactly one of the inequalities listed in the previous paragraph.
 
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Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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