341For maximum value we choose x,y,z such that ax+by+cz>0
(a-x)(b-y)(c-z)(ax+by+cz) = \frac{1}{abc} (a^2-ax)(b^2-by)(c^2-cz)(ax+by+cz)
\le \frac{1}{abc} \left(\frac{(a^2-ax)+(b^2-by)+(c^2-cz)+(ax+by+cz)}{4}\right)^4
=\frac{1}{abc} \left(\frac{a^2+b^2+c^2}{4}\right)^4
Equality occurs when a-x = b-y = c-z = ax+by+cz
Now we have to examine that our conditions for equality do not contradict our requirements that the factors are each positive
So let a-x = b-y = c-z = ax+by+cz = k. So we need k>0
Its easy to see that from the above equations that k = \frac{a^2+b^2+c^2}{1+a+b+c}>0
