Arbitrarily give four positive numbers, assuming $A_1, B_1, C_1, D_1$; In $A_1$ and $B_1$, subtract the smaller from the larger (0 if equal), and the result is $A_2$; In $B_1$ and $C_1$, subtract the smaller from the larger (0 if equal), and the result is $B_2$; In $C_1$ and $D_1$, subtract the smaller from the larger (0 if equal), and the result is $C_2$; In $D_1$ and $A_1$, subtract the smaller from the larger (0 if equal), and the result is $D_2$;
After a round of calculation, $A_2, B_2, C_2$, and $D_2$ will be obtained from the original $A_1, B_1, C_1$, and $D_1$; If we recalculate from $A_2, B_2, C_2$, and $D_2$, we obtain $A_3, B_3, C_3$, and $D_3$; After multiple rounds of calculation, what results will be obtained?
Example: If four numbers are first given as $15, 8, 2, 12$; Then $15-8$ gets $7$, $8-2$ gets $6$, $12-2$ gets $10$, $15-12$ gets $3$, and four new numbers are obtained; Continuing with this calculation, we find that the last four numbers are equal, all equal to $3$.
My idea is to use $\max(x,y)-\min(x,y)=|x-y|$.
$\max(|a-b|,|b-c|,|c-d|,|d-a|)=\max(a-b,b-c,c-d,d-a)$.
So we need to prove that
$\max (a-b,b-c,c-d,d-a)-\min(a-b,b-c,c-d,d-a)\leq \max(a,b,c,d)-\min(a,b,c,d)$.
I can only prove that $\max{a-b,b-c,c-d,d-a}\leq\max{a,b,c,d}$. Therefore, after each transformation, the maximum value of the four numbers decreases (sometimes remains unchanged), and the four numbers eventually become smaller and smaller until they are equal.
I do not know how to proceed next. Any help?