# Inequality proof

## Homework Statement

Prove that if x and y are not both 0 then x2 + xy + y2 > 0

## 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 x2 + xy + y2
= x2 + x2 + x2
= y2 + y2 + y2 = x2 + xy + y2 > 0 since the square of a real number is
always greater than or equal to zero and x= y $$\neq$$ 0.
Case 2. x > y. Then (x - y) > 0 and x3 - y3 > 0.
Therefore, x3 - y3 = x3 + 0 - y3
= x3 + (x2y - x2y + xy2 - xy2) - y3 = ( x - y)( x2 + xy + y2) > 0.
Dividing both sides by ( x - y) we get x2 + xy + y2 which is greater
than zero since x and y are not both zero.
Case 3. x < y. Then y - x > 0 and y3 - x3 > 0. Therefore,
y3 - x3 = y3 + 0 - x3
= y3 + ((x2y - x2y + xy2 - xy2) - x3 = (y - x)( x2 + xy + y2) > 0.
Dividing both sides by ( y - x) we get x2 + xy + y2 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

## The Attempt at a Solution

Related Precalculus Mathematics Homework Help News on Phys.org
tiny-tim
Homework Helper
Hi nike5!
Prove that if x and y are not both 0 then x2 + xy + y2 > 0

## 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 …
eugh! :yuck:

you mustn't try to solve everything by the trichotomy law!

Hint: complete the square

ahhhh I feel like an idiot for not seeing that.

Proof. Let x and y be arbitrary real numbers. Suppose x and y do not both equal 0.
Then x2 + xy + y2 = (x2 + xy) + y2
= ( x + (y / 2))2 + (3y2) / 4. Therefore,
( x + (y / 2))2 $$\geq$$ 0 and (3y2) / 4 $$\geq$$ 0 since the square of any real number is positive. Under the closure property of addition,
( x + (y / 2))2 + (3y2) / 4 $$\geq$$ 0 and since x and y are not both zero we can conclude that ( x + (y / 2))2 + (3y2) / 4
= x2 + xy + y2 > 0.

tiny-tim