What is the difference between the largest and smallest possible values of any of these numbers? 
If $x,y,z$ are real numbers such that $x+y+z = 5$ and $xy+yz+zx = 3$, what is the difference between the largest and smallest possible values of any of these numbers?

What is wrong with this approach?
We have that $(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+xz) = 25 \implies x^2+y^2+z^2 = 19$. Thus the points $(x,y,z)$ lie on a sphere with radius $\sqrt{19}$. Therefore the maximum value of any of these is $\sqrt{19}$ and the smallest is $-\sqrt{19}$ and the difference is $2\sqrt{19}$.
 A: $(x,y,z)$ is not $\bf{three}$ points. It is a single point in three dimensional space, where $x,y,z$ are the coordinates.
Now think of a sphere of radius $\sqrt{19}$. If the $x$ coordinate is $\sqrt{19}$, then is it possible for any other coordinate to be $-\sqrt{19}$?
A: The problem with your approach is that the point $(x,y,z)$ is not only on the sphere $x^2+y^2+z^2=19$, it is also on the plane $x+y+z=5$.  That is, it is on the intersection of this sphere and plane, which is a circle.  You will need to find the maximum and minimum coordinates of points on this circle.
As far as I can see this problem is quite tricky to visualise geometrically.  It can be solved by Lagrange multipliers (though since you tagged the question algebra-precalculus that is probably not the method you are expected to use).  If my calculations are correct the maximum possible difference is $8/\sqrt3$.
A: You have shown that the points are on the sphere you say, but it does not establish they can be anywhere on the sphere. If I try to plug $x=\sqrt{19}$ into the equations, $y$ and $z$ come out complex, so $x$ cannot be that large.
