complete the square $ y^2 + z^2 + yz - 5y - 5z $ and showing it is bounded

Question

I want to show that the expression : $$ y^2 + z^2 + yz - 5y - 5z = -8$$ is bounded. I was thinking about writing the expression in the form of squares, just like here:

$$ 1= x^2 + y^2 + xy - 3x - 3y = [x-1 + \frac 1 2 (y-1) ]^2 + \frac 3 4 (y-1)^2 $$

do you have any idea how to do this ? If you have any other method, I'm also taking it.

Answer 1

You can start by putting the $yz$ term inside a square and write $$y^2 + z^2 + yz - 5y - 5z = {(y + z)^2 \over 2} + {y^2 \over 2} + {z^2 \over 2} - 5y - 5z$$ Next, put the $5y$ and $5z$ terms inside squares: $${(y + z)^2 \over 2} + {y^2 \over 2} + {z^2 \over 2} - 5y - 5z = {(y + z)^2 \over 2} + {(y - 5)^2 \over 2} + {(z-5)^2 \over 2} - 25$$ So your points are the $(y,z)$ for which $${(y + z)^2 \over 2} + {(y - 5)^2 \over 2} + {(z-5)^2 \over 2} = 17$$ So in particular, ${(y - 5)^2 \over 2}$ and ${(z-5)^2 \over 2}$ are both at most $17$. So $|y- 5|, |z - 5| \leq \sqrt{34}$ and your set is bounded.

Answer 2

You can do it like this.

$$y^2+z^2+yz-5y-5z+8=0$$

Let's first deal with all the terms in $y$. Note that $(y+\frac z2)^2$ has a term $yz$ and $(y-\frac 52)^2$ has a term $-5y$. To eliminate the fractions first multiply by $4$.

$$4y^2+4z^2+4yz-20y-20z+32=0$$

Then, taking the two clues about eliminating $y$ we take a summand $$(2y+z-5)^2=4y^2+4yz-20y+z^2-10z+25$$ so that $$(2y+z-5)^2+3z^2-10z+7=0$$

If we now multiply by $3$ we obtain $$3(2y+z-5)^2+9z^2-30z+21=0$$ or, completing the square in the $z$ terms: $$3(2y+z-5)^2+(3z-5)^2=4$$

Which may give you what you want - this bounds $z$ and by symmetry $y$ can be bounded too.

Answer 3

The graph of $ y^2 + z^2 + yz - 5y - 5z = -8$ is clearly symmetric about the line $y=z$.

To get it into a more familiar form, apply the following transformation (rotate $45^o)$:

$x=\dfrac y{\sqrt2}-\dfrac z{\sqrt2}; \; w=\dfrac y{\sqrt2}+\dfrac z{\sqrt2}.$

The equation then becomes $x^2+w^2+\dfrac{w^2-x^2}2-5\sqrt2w=-8$ or $x^2+3w^2-10\sqrt2w=-16.$

Now complete the square to get $x^2+3\left(w-\dfrac{5\sqrt2}3\right)^2=-16+\dfrac{50}3=\dfrac23.$

This is a familiar form for an ellipse.

In any event, from this equation it is clear that $x$ and $w$ -- and therefore $y$ and $z$ -- are bounded.