finding x,y,z for optimizing expression

Question

The question is as follows:

let (x,y,z) be an ordered triplet of real numbers such that x<1 ,y<2 ,z<3 and x+ y/2 + z/3 >0.

for x=a ,y=b,z=c the value of expression (1-x)(2-y)(3-z)(x+y/2+z/3) is maximum and we need to find a+b+c.

my approach:

since the terms in the product are positive ,hence we can use AM-GM inequality and hence product is maximum when

1-x = 2-y = 3-z = x+y/2+z/3 = t (say)

solving for t gives t=18/17 which clearly makes x<0 which is not allowed.

hence I am stuck.kindly help.

Answer 1

Use AM-GM for the following: $$6(1-x)\left(1-\frac{y}{2}\right)\left(1-\frac{z}{3}\right)\left(x+\frac{y}{2}+\frac{z}{3}\right).$$ The equality occurs for $$1-x=1-\frac{y}{2}=1-\frac{z}{3}=x+\frac{y}{2}+\frac{z}{3},$$ which gives values of $a$, $b$ and $c$.

Can you end it now?

I got $a+b+c=\frac{3}{2}.$

Answer 2

change variables to $u = y/2$ and $v = z/3$ so $0 \le u,v \le 1$ and you need to find $a+2u+3v$ where we maximize $$ (1-x)(2-y)(3-z)(x+y/2+z/3) = 6(1-x)(1-u)(1-v)(x+u+v). $$ So we must have $x=u=v$ and $1-x = x+u+v = 3x \iff x = u = v = 1/4$.

Can you finish?