Comparison Proof via axioms, almost done need hints for finish and proof read

[silvermane]

1. The problem statement:

Prove that

0\leq2x^2 - 3xy + 2y^2​

2. These are the axioms we are permitted to use:

01) Exactly one of these hold: a<b, a=b, or b<a
02) If a<b, and b<c, then a<c
03) If a<b, then a+c < b+c for every c
04) If a<b and 0<c, then ac<bc.

The Attempt at a Solution

By axiom 03) we are permitted to add +-y^2 to both sides:
0 = -y^2 + y^2 \leq 2x^2 -3xy+2y^2-y^2+y^2
0 \leq 2x^2 -3xy +y^2 + y^2

Again, by axiom 03) we add +-x^2 to both sides:
0 = x^2 - x^2 \leq 2x^2 - 3xy +y^2 +y^2 + x^2 - x^2 = x^2 - 3xy + y^2 +y^2 + x^2

By axiom 03) again, we add +-xy to both sides:
0 = xy - xy \leq 2x^2 - 3xy +y^2 +y^2 + x^2 - x^2 +xy - xy = x^2 - 3xy + y^2 +y^2 +xy +x^2

Case 1: If x=y=0, then our expression is zero

Case 2: If x doesn't = y, and y=0, then we have 0&lt;x^2+x^2=2x^2

Case 3: If x doesn't = y, and x=0, then we have 0&lt;y^2

Case 4: If x doesn't = y, and y\neq0,
0 &lt; (x-y)^2 &lt; (x-y)^2 +x^2 +xy +y^2 ??
then I'm stuck and need help

If anyone can help me figure out the last part, it would be greatly appreciated. I've worked very hard trying to figure it out, but that xy we would get could possibly be negative and I just need to prove that last part.

Thank you in advance for any hints/tips!

 

[╔(σ_σ)╝]

Case 1

x,y &gt;0 This is trivial since everything is positive.

Case 2
x &gt;0 ,y &lt;0 \Rightarrow x&gt;y

-x^{2}&lt; xy \Rightarrow 0&lt; xy + x^{2}

(x-y)^{2} + y^{2}&lt; (x-y)^{2} +x^{2} + xy +y^{2}


Case 3

This is similar to case 2.


How come you can't use the distributive properties ?

Without them what does xy mean ? And how does it differ from yx ?


Assuming you could use axioms of multiplication and the distributive property you could divide by 2 and complete the square and the result falls into your hands.

x^2 - \frac{3}{2}xy + 2y^2 = (x- \frac{3}{2}y)^{2} + \frac{7}{16}y^{2}