Prove that $(a-b)(b-c)(a-c) \leq 2$ [ when $0 \leq a \leq 2$ , $0 \leq b \leq 2$ , $0 \leq c \leq 2$ ]

Question

Let $a$ , $b$ and $c$ be real numbers such that $0 \leq a \leq 2$ , $0 \leq b \leq 2$ , $0 \leq c \leq2$ .

Prove that $(a-b)(b-c)(a-c) \leq 2$

When does that equality hold?

Answer 1

Let's denote $L:= (a-b)(b-c)(c-a)$.

WOLG, there are only two cases: $a\le b \le c$ or $a\ge b \ge c$.

In the first case, the term $L$ is negative (then, it is smaller than $2$), so, we focus only on the second case where $a\ge b \ge c$. Applying the AM-GM inequality: $$\begin{align} L&= 2 (a-b)(b-c)\left(\frac{a-c}{2} \right)\\ \le &2\cdot \left(\frac{(a-b)+(b-c)+\left(\frac{a-c}{2} \right)}{3} \right)^3= \frac{(a-c)^3}{4} \le \frac{(2-0)^3}{4} = 2 \end{align}$$

The equality occurs if and only if $(a,b,c) = (2,1,0)$.

Answer 2

Consider the number line , where we have $0,1,2$.
Somewhere in that interval , we have the variables $A,B,C$ in some order.

Let us reorder (rename) the variables such that we have the order $A,B,C$
We want to maximize $X(A,B,C)=|(A-B)(B-C)(C-A)|$

When $|(A-C)|$ ( which is the largest term here ) is kept constant , then it is easy to see that $X$ is locally maximized when $B$ is in the Exact Middle , that is $|(A-B)|=|(B-C)|$.

We can increase this local maxima by increasing $|(A-C)|$
Increasing $|(A-C)|$ will give us $|(A-C)|=2$ , where $A=0$ & $C=2$

The global maxima will then occur when $B$ is in the Exact Middle , that is $B=1$.
Hence , Maximum $X$ is $|(0-2)(2-1)(1-0)|=2$

All other $X$ values will be less than that maximum.
Equality occurs when-ever the variables are $(0,1,2)$ in some order.