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 $x&lt;1$, $y&lt;2$, $z&lt;3$, and $x+\frac{y}{2}+\frac{z}{3}&gt;0$. 


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.
$$6(1-x)(1-y/2)(1-z/3)(x+y/2+z/3) \leq 6\Big({ (1-x) +...\ \over 4}\Big)^4 =6\cdot 3^4/4^4$$
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 $x= 1/4$...
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.