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I want to prove the following identity: For every real numbers $a,b,c,$ \begin{gather*} \min\{\max\{a,b\},\max\{a,c\},\max\{b,c\}\}=\max\{\min\{a,b\},\min\{a,c\},\min\{b,c\}\}. \end{gather*} Clearly, let \begin{gather*} L(a,b,c):=\min\{\max\{a,b\},\max\{b,c\},\max\{c,a\}\}, \\ R(a,b,c):=\max\{\min\{a,b\}, \min\{b,c\},\min\{c,a\}\}. \end{gather*} It is clear that $L$ and $R$ are cyclically symmetric. Thus, it is easy to prove $L(a,b,c)=R(a,b,c),$ by just considering the case that $a\geq b\geq c.$ But my question is: Can it possible to prove the statement, without using cyclic symmetry?

Actually, I have finite proving the first part, as follows: Because \begin{gather*} \min\{a,b\}\leq \max\{a,b\},\quad \min\{b,c\}\leq \max\{a,b\},\quad \min\{c,a\}\leq \max\{a,b\}, \end{gather*} we see that $\max\{a,b\}$ is an upper bound of the set $$\{\min\{a,b\},\min\{b,c\},\min\{c,a\}\},$$ and so, $$\max\{\min\{a,b\},\min\{b,c\},\min\{c,a\}\}\leq \max\{a,b\}.$$ Similarly, from \begin{gather*} \min\{a,b\}\leq \max\{b,c\},\quad \min\{b,c\}\leq \max\{b,c\},\quad \min\{c,a\}\leq \max\{b,c\} \end{gather*} we deduce that $$\max\{\min\{a,b\},\min\{b,c\},\min\{c,a\}\}\leq \max\{b,c\},$$ and from \begin{gather*} \min\{a,b\}\leq \max\{c,a\},\quad \min\{b,c\}\leq \max\{c,a\},\quad \min\{c,a\}\leq \max\{c,a\}, \end{gather*} we see that $$\max\{\min\{a,b\},\min\{b,c\},\min\{c,a\}\}\leq \max\{c,a\}.$$ From above we deduce that the number $$\max\{\min\{a,b\},\min\{b,c\},\min\{c,a\}\}$$ is a lower bound of the set $$\{\max\{a,b\},\max\{b,c\},\max\{c,a\}\},$$ and so $$\max\{\min\{a,b\},\min\{b,c\},\min\{c,a\}\}\leq \min\{\max\{a,b\},\max\{b,c\},\max\{c,a\}\}.$$

But how to prove the converse inequality, that is, $$\min\{\max\{a,b\},\max\{b,c\},\max\{c,a\}\}\leq \max\{\min\{a,b\},\min\{b,c\},\min\{c,a\}\}?$$

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azc
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    possible to prove the statement, without using cyclic symmetry? You could write down the individual cases of sort ordering separately and verify the equality for each. Same idea, just more work. – dxiv Feb 13 '18 at 06:34
  • @dxiv: Yes, it is certainly OK to do so. But I do not want to prove by cases! – azc Feb 13 '18 at 06:42
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    Maybe you should then explain better why you don't want to use the symmetry, either. – dxiv Feb 13 '18 at 06:45
  • @dxiv: I just want to find alternative proof. BTW, with the result, I shall prove the following identity concerning gcd and lcm: $([a,b],[b,c],[c,a])=[(a,b),(a,c),(b,c)].$ – azc Feb 13 '18 at 07:11
  • It is not correct to assume that $a\geq b\geq c$ if you only have cyclic symmetry; you need full symmetry in the three variables (i.e., symmetry for the entire symmetric group $S_3$, not just a cyclic subgroup). – Eric Wofsey Feb 13 '18 at 07:30
  • @EricWofsey: You are right! Since I think transpositional symmetry is apparent, so I do not mention it. – azc Feb 13 '18 at 07:51
  • I thought for a moment you could simply apply the inequality you proved to $-a, -b, -c$ and use the fact that $\max(x,y,z)=-\min(-x,-y,-z)$. Unfortunately I think that flips the inequality again so you end up with the same thing. – Jaap Scherphuis Feb 13 '18 at 13:20
  • Related: https://math.stackexchange.com/questions/3499913/what-divisibility-function-is-between-gcd-and-lcm-with-three-inputs – mr_e_man Nov 06 '23 at 20:26

1 Answers1

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Let us make a slight generalization, considering not just three variables but $n$ variables where $n\geq 3$. So, let $x=(x_1,x_2,\ldots,x_n)$ be an uple of real numbers. Define

$$l_k(x)=\max(x_j,j\neq k), r_k(x)=\min(x_j,j\neq k) \tag{1}$$ and $$L(x)=\min_{1\leq k \leq n}l_k(x), R(x)=\max_{1\leq k \leq n} r_k(x) \tag{2}$$

There is a permutation $\sigma$ of $\lbrace 1,2,3,\ldots,n \rbrace$ such that $x_{\sigma(1)} \leq x_{\sigma(2)} \leq \ldots \leq x_{\sigma(n)}$. Put $y_k=x_{\sigma(k)}$ for $1 \leq k \leq n$. Then

$$y_1 \leq y_2 \leq \ldots \leq y_n \tag{3}$$

Since $l_k(x)=l_{\sigma(k)}(y)$, the sets $\lbrace l_k(x) | 1\leq k \leq n\rbrace$ and $\lbrace l_k(y) | 1\leq k \leq n\rbrace$ coincide. We deduce $L(x)=L(y)$, and similarly $R(x)=R(y)$.

Now, by (3), we have $l_n(y)=y_{n-1}$ and $l_k(y)=y_n$ for $k<n$, so $L(y)=y_{n-1}$. Similary $R(y)=y_2$. Finally,

$$ L(x)=y_{n-1}, R(x)=y_2 \tag{4} $$

When $n=3$, the two coincide.

Ewan Delanoy
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