Determine the largest and smallest of the rational numbers $x$, $y$, $z$ if $x^2+4y^2+z^2-4xy-4yz+2xz-2x+2z+2=0$

Question

Determine what number out of three rational numbers $x$, $y$, $z$ is the biggest and smallest one if following equality is true: $$x^2+4y^2+z^2-4xy-4yz+2xz-2x+2z+2=0$$

I think it may be somewhat connected with full squares extraction

$$(x-2y)^2+(z+1)^2-2z(2y-x)-2x+1=0$$

Or

$$(z-2y)^2+(x-1)^2-2x(2y-z)+2z+1=0$$

Or

$$(x+z)^2-4y(y-x-z)-2(x-z-1)=0$$

But i am still not quite sure, what to do next

Answer 1

Let's start by rewriting the equality as

$$ (x+z-2y)^2 + 2 = 2(x-z) $$

Now complete the square on the LHS first by adding $2(x+z-2y)$ to both sides: $$ (x+z-2y+1)^2 + 1 = 2(x-z) + 2(x+z-2y) = 4(x-y) $$ And again complete the square by subtracting $2(x+z-2y)$: $$ (x+z-2y-1)^2 + 1 = 2(x-z) - 2(x+z-2y) = 4(y-z) $$ Since LHS is positive in all 3 equations, you can now determine the ordering between the 3 variables: $x > y >z$

Answer 2

There are a couple of ways to get squared terms (I can think of three) from the above equation. If you use the fact the any squared term is greater than zero, then the rest of the terms are less than zero. If you then work with the generated inequalities, you may be able to order the given variables.