Why does the equality $\max\{a,\min\{b,c\}\}=\min\{\max\{a,b\},\max\{a,c\}\}$ hold? 
Let $a,b,c$ be three real numbers. Show that the following equality holds.
$$\max\{a,\min\{b,c\}\}=\min\{\max\{a,b\},\max\{a,c\}\}$$

 A: Since the formulas are invariant  with respect to the swapping of $b$ and $c$, consider $b\leq c$ and the three cases: 
$$\mbox{i) }a\leq b\leq c,\quad  
\mbox{ii) } b\leq a\leq c, \quad 
\mbox{iii) } b\leq c\leq a.$$
Then 
i) $\max\{a,\min\{b,c\}\}=\max\{a,b\}=b$, and  $\min\{\max\{a,b\},\max\{a,c\}\}=\min\{b,c\}=b$;
ii) $\max\{a,\min\{b,c\}\}=\max\{a,b\}=a$, and  $\min\{\max\{a,b\},\max\{a,c\}\}=\min\{a,c\}=a$;
iii) $\max\{a,\min\{b,c\}\}=\max\{a,b\}=a$, and  $\min\{\max\{a,b\},\max\{a,c\}\}=\min\{a,a\}=a$;
and we are done.
A: Consider the sets $S_A=(-\infty,A)$, $S_B=(-\infty,B)$ and $S_C=(-\infty,C)$. See that:
$$
(-\infty,A)\cap (-\infty,B)=(-\infty,\min\{A,B\}),\\
(-\infty,A)\cup (-\infty,B)=(-\infty,\max\{A,B\}).
$$
So your problem is equivalent to:
$$
S_A\cap(S_B\cup S_C)=(S_A\cup S_B)\cap(S_A\cup S_C),
$$
which is an elementary set theoretic fact.
A: First of all note that the equation still holds if one adds a number $t$ to each of $a, b$ and $c$, so lets add $t = -a$. Then we can use the identies $max(x,y) = \frac{x + y + \left\lvert{x - y}\right\rvert}{2}$ and $min(x,y) = \frac{x + y - \left\lvert{x - y}\right\rvert}{2}$. Substituting this into the given equations gives the identity:
$$b + c - \left\lvert{b - c}\right\rvert + \left\lvert{b  + c  - \left\lvert{b - c}\right\rvert}\right\rvert = b + c + \left\lvert{b}\right\rvert + \left\lvert{c}\right\rvert - \left\lvert{b - c + \left\lvert{b}\right\rvert - \left\lvert{c}\right\rvert}\right\rvert$$
A: In a similar spirit to Arash's answer:
If we define real numbers using dedekind cuts, then as sets, $\max\{a,b\}=a\cup b$ and $\min\{a,b\}=a\cap b$.
The equation then becomes 
$$a\cap (b\cup c)=(a\cap b)\cup (a\cap c)$$
Which is easy to prove.
