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, $$\min\{\max\{a,b\},\max\{b,c\},\max\{c,a\}\}\leq \max\{\min\{a,b\},\min\{b,c\},\min\{c,a\}\}?$$
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.