If $x,y,z>0$ and $x+y+z=1$,then find the maximum value of $(1-x)(2-y)(3-z)$.

Question

If $x,y,z>0$ and $x+y+z=1$,then find the maximum value of $(1-x)(2-y)(3-z)$.

My Attempt:

We have $0<1-x<1,1<2-y<2,2<3-z<3$ so A.M-G.M inequality cannot be used since we can never have

$1-x=2-y=3-z$.

Is there any other way out

Answer 1

Let us put $z=1-x-y$. Then the expression becomes

$$(1-x)(2-y)(2+x+y).$$

Rewrite it as:

$$x^2\cdot(y-2)+x\cdot(y^2-y-2)+(4-y^2).$$

If we consider this as a quadratic function of $x$ then we see that the branches of parabola look down, and its vertex is at

$$-\frac{y^2-y-2}{2(y-2)}=-\frac{y+1}2.$$

As $y$ changes from $0$ to $1$, $-\frac{y+1}2$ changes from $-1/2$ to $-1$. Hence the maximum is when $x=0$.

Our expression then becomes $$(2-y)(2+y),$$ which has supremum at $4$ when $y\to 0$.

Answer 2

As pointed out in the comments, equality holds at a boundary point. With that in mind, we use a smoothing argument, and show that

$$f(x,y,z) \leq f(0, y, x+z) \leq f(0, 0, x+y+z) = f(0, 0, 1) = 4.$$

Each of the inequalities should be obvious.

1. Holding $x+z$ constant, $(1-x) \times (3-z)$ is maximized when the terms are as close to $\frac{4-x-y}{2}$ as possible.
   • We can't set $1-x = 3-z$ as that requires $ 1 \geq z-x = 3-1 = 2$. (Notice that the AM-GM attempts fail at this step, because they don't consider the boundary condition.)
   • This is where we run into the boundary condition of $ x \geq 0, z \leq 1$. The expression is maximal when $x^* = 0, z^* = x+z$.
2. Likewise, holding $y+z$ constant, then $(2-y) \times (3-z)$ is maximized when the terms are as close together as possible.
   • As before, the expression is maximal when $y^* = 0, z^* = y+z$.

Notes

• Aig's solution is similar. Expressed along these lines, they are saying that when we hold $y$ constant (which is equivalent to holding $x+z$ constant), then the value is maximized at $ x^* = 0$. After that, holding $x = 0$ constant, then the value is maximized at $y^* = 0 $.
• Since the supremum occurs at a boundary condition of $(0, 0, 1)$, AM-GM doesn't easily work because it's hard to conjure up an equality condition to occur exactly at the boundary condition.

Answer 3

Remarks: Once we motivate the point $(0, 0, 1)$, we can write a terse solution. Often, the motivation of a solution is complicated while the solution itself is terse.

We have \begin{align*} (1-x)(2-y)(3-z) &= (2 -2x - y + xy)(3 - z)\\ &\le (2 - 2x - y + x)(3 - z)\\ &= (2 - x - y)(3 - z)\\ &= (1 + z)(3 - z)\\ &= -(z - 1)^2 + 4\\ &\le 4. \end{align*}