Find Maximum Value of Expression I'm trying to learn about using the AM-GM inequality. I understand the basics of it so far, but I'm having a difficult time in terms of applying it to solve problems. I've encountered the problem below and just can't seem to find a way to solve it.

Find the maximum value of the expression $$(1-x)(2-y)(3-z)(x+\frac{y}{2}+\frac{z}{3})$$
  given that $x<1$, $y<2$, $z<3$, and $x+\frac{y}{2}+\frac{z}{3}>0$. 

So, how do you? It seems that expanding the expression above just makes it more complicated. Should I even try to use AM-GM for this?
Thanks in advance.
 A: $$6(1-x)(1-y/2)(1-z/3)(x+y/2+z/3) \leq 6\Big({ (1-x) +...\ \over 4}\Big)^4 =6\cdot 3^4/4^4$$
with equality iff $1-x = 1-y/2 = 1-z/3 = x+y/2+z/3$
So iff $ y=2x$ and $z=3x$ we get $x= 1/4$...
A: Consider $a(1-x)+b(2-y)+c(3-z)+d\left(x+\dfrac y2+\dfrac z3\right)$
$=x(d-a)+y\left(\dfrac d2-b\right)+z\left(\dfrac d3-c\right)+a+2b+3c$
Set the coefficients of $x,y,z$ to $=0$
$\implies a=d,b=\dfrac d2,c=\dfrac d3$
WLOG $d=[1,2,3]=6$
Using GM $\le AM$
$$\sqrt[6+3+2+6]{6(1-x)\cdot3(2-y)\cdot2(3-z)\cdot6\left(x+\dfrac y2+\dfrac z3\right)}$$
$$\le\dfrac{6(1-x)+3(2-y)+2(3-z)+6\left(x+\dfrac y2+\dfrac z3\right)}{6+3+2+6}=\dfrac{24}{23}$$
A: Rewrite $$(1-x)(2-y)(3-z)(x+\frac{y}{2}+\frac{z}{3})=6(1-x)(2-y)(3-z)(x-1+\frac{y}{2}-1+\frac{z}{3}-1+3)$$
and set $\;a=1-x,\; b=1-\frac{y}{2},\;c=1-\frac{z}{3}\;$ to get
$$(1-x)(2-y)(3-z)(x+\frac{y}{2}+\frac{z}{3})=6abc(3-a-b-c)$$
With the use of AM-GM is
$$6abc(3-a-b-c)\leq 6\left(\frac{a+b+c+3-a-b-c}{4}\right)^4$$
which is easy to finish.
A: By AM-GM $$(1-x)(2-y)(3-z)\left(x+\frac{y}{2}+\frac{z}{3}\right)=6(1-x)\left(1-\frac{y}{2}\right)\left(1-\frac{z}{3}\right)\left(x+\frac{y}{2}+\frac{z}{3}\right)\leq$$
$$\leq6\left(\frac{1-x+1-\frac{y}{2}+1-\frac{z}{3}+x+\frac{y}{2}+\frac{z}{3}}{4}\right)^4=\frac{243}{128}.$$
The equality occurs for $$1-x=1-\frac{y}{2}=1-\frac{z}{3}=x+\frac{y}{2}+\frac{z}{3},$$ which says that we got a maximal value. 
