Proving Inequality: x^2 + xy + y^2 > 0 for Non-Zero x and y

[nike5]

Homework Statement

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

 

[tiny-tim]

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 ​

 

[nike5]

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]

Woohoo! ​