IMHO, the method in the link is a fallacy, or at least, it's a source of misunderstanding. The wrong step is that from this equation:
[tex]v_{rel}=\sqrt{\vec{v_1}^2-2\vec{v_1}\vec{v_2}+\vec{v_2}^2[/tex]
we CANNOT go straight to this equation:
[tex]<v_{rel}>=\sqrt{<\vec{v_1}^2>-<2\vec{v_1}\vec{v_2}>+<\vec{v_2}^2>[/tex]
simply because these two are different:
[tex]<\sqrt{\vec{v_1}^2-2\vec{v_1}\vec{v_2}+\vec{v_2}^2}>\neq \sqrt{<\vec{v_1}^2>-<2\vec{v_1}\vec{v_2}>+<\vec{v_2}^2>[/tex]
That this method yields the correct result, I think, is just a mathematical coincidence.
We have the relative speed between 2 particles: [tex]v_{rel}=|\vec{v_1}-\vec{v_2}| = 2v|cos\phi |[/tex] where [tex]\phi[/tex] is one-half of the angle occupied by 2 vectors [tex]\vec{v_1}[/tex] and [tex]-\vec{v_2}[/tex]. Now some questions:
1. Because of the randomness of the gas, in the viewpoint of each molecule, the average speed of all other molecules relative to it (let V denote this average speed) is the same for every molecule; that is, the average speed in the viewpoint of each molecule doesn't depend on which molecule considered. Do you agree?
2. From that, what can you deduce about the relation between the average relative speed of the whole system and V? Do we only need to compute V in order to calculate the needed average relative speed?
3. Calculating V: Pick an arbitrary molecule with velocity [tex]\vec{u}[/tex]. You can use [tex]\vec{u}[/tex] as a fixed axis and calculate [tex]<|cos\phi |>[/tex]. Notice the uniform distribution of velocities in every direction, which leads to spherical symmetry.
P.S.: The < > sign means average.