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## Homework Statement

Prove that if x and y are not both 0 then x

^{2}+ xy + y

^{2}> 0

## Homework Equations

## The Attempt at a Solution

Proof. Let x and y be arbitrary real numbers. Suppose x and y are not both zero. We will consider three cases.

Case 1. x = y. Then x

^{2}+ xy + y

^{2}

= x

^{2}+ x

^{2}+ x

^{2}

= y

^{2}+ y

^{2}+ y

^{2}= x

^{2}+ xy + y

^{2}> 0 since the square of a real number is

always greater than or equal to zero and x= y [tex]\neq[/tex] 0.

Case 2. x > y. Then (x - y) > 0 and x

^{3}- y

^{3}> 0.

Therefore, x

^{3}- y

^{3}= x

^{3}+ 0 - y

^{3}

= x

^{3}+ (x

^{2}y - x

^{2}y + xy

^{2}- xy

^{2}) - y

^{3}= ( x - y)( x

^{2}+ xy + y

^{2}) > 0.

Dividing both sides by ( x - y) we get x

^{2}+ xy + y

^{2}which is greater

than zero since x and y are not both zero.

Case 3. x < y. Then y - x > 0 and y

^{3}- x

^{3}> 0. Therefore,

y

^{3}- x

^{3}= y

^{3}+ 0 - x

^{3}

= y

^{3}+ ((x

^{2}y - x

^{2}y + xy

^{2}- xy

^{2}) - x

^{3}= (y - x)( x

^{2}+ xy + y

^{2}) > 0.

Dividing both sides by ( y - x) we get x

^{2}+ xy + y

^{2}which is greater than zero since x and y are not both zero.

Truth?

I greatly appreciate anyone's help. I'm studying independently and don't have anyone to check my proofs