- #1

silvermane

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**1. The problem statement:**

Prove that

[tex]0\leq2x^2 - 3xy + 2y^2[/tex]

**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 [tex]+-y^2[/tex] to both sides:

[tex]0 = -y^2 + y^2 \leq 2x^2 -3xy+2y^2-y^2+y^2[/tex]

[tex]0 \leq 2x^2 -3xy +y^2 + y^2[/tex]

Again, by axiom 03) we add [tex]+-x^2[/tex] to both sides:

[tex]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[/tex]

By axiom 03) again, we add +-xy to both sides:

[tex]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[/tex]

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

Case 2: If x doesn't = y, and y=0, then we have [tex]0<x^2+x^2=2x^2[/tex]

Case 3: If x doesn't = y, and x=0, then we have [tex]0<y^2[/tex]

Case 4: If x doesn't = y, and [tex]y\neq0[/tex],

[tex]0 < (x-y)^2 < (x-y)^2 +x^2 +xy +y^2 [/tex] ??

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!