How to prove this simple property for two sets? We are given two vectors, $a = (a_1,\dots,a_n)$ and $b = (b_1,\dots,b_n)$ such that $0 \le a_i=b_i \le \varDelta$ for $i=1,\dots,n$. We want to modify each of these vectors in an iterative procedure such that in each iteration one component of each vector is increased. However, we do this for each vector differently.
Let $0 \le \delta \le \varDelta$ be a given number. 
For vector $a$, we select the minimum component, say $a_k$, and increase it by $\min(\delta,\varDelta-a_k)$. For vector $b$, we select an arbitrary component, say $b_k$, and increase it by $\min(\delta,\varDelta-b_k)$. Let $a^\ell$ and $b^\ell$ represent the modified vectors at iteration $\ell$. At the beginning, iteration $0$, we have $a^0 = b^0=a=b$.
It is quite clear that
$$
\min \{a^\ell_1,\dots,a^\ell_n\} \ge \min  \{b^\ell_1,\dots,b^\ell_n\}
$$
for any iteration $\ell$. However, I have difficulty proving this simple logical fact, mathematically! I tried induction, proved it for $\ell=1$, but could not succeed proving the next steps. Given that our statement is logically trivial, there should be some easy way to prove this. Any ideas? If that helps, all parameters are non-negative integers.
 A: We can ignore the saturation role of $\Delta$ and can wlog assume that $\delta=1$.
By introducing infinitesimal changes to the original $a_i$, we may assume that $a_i-a_j$ is never an integer for $i\ne j$. As a consequence,  $\min A^\ell$ is a strictly increasing function of $\ell$. 
Note that $$ a_i^\ell=a_i+c_i^\ell$$
with $c_i^\ell\in\Bbb N_0$. 
We have


*

*If $c_i^\ell=0$, then $\min A^\ell\le a_i$ (obviously)

*If $c_i^\ell >0$, then $a_i+c_i^\ell-1< \min A^\ell\le a_i+c_i^\ell$: The right inequality is trivial, if the left failed, we would not have $a_i+c_i^{\ell'}\le \min A^{\ell'}$ at the step $\ell'$ where $c_i$ was last increased.


(In other words, $c_i^\ell = \min\bigl\{0,\lceil \min A^\ell-a_i\rceil\bigr\}.$)
We also have $b_i^\ell = a_i+d_i^\ell$ with $d_i^\ell\in \Bbb N_0$ and with $\sum c_i^\ell=\sum d_i^\ell = \ell$. 
If $d_i^\ell \ge c_i^\ell$ for all $i$, then from the equality of sums, $c_i^\ell=d_i^\ell$ for all $i$, so $A^\ell=B^\ell$ and trivially $\min A^\ell\ge \min B^\ell$. Hence we can assume that $d_i^\ell< c_i^\ell$ for some $i$.
Then
$$ \min B^\ell\le a_i+d_i^\ell\le a_i+c_i^\ell-1<\min A^\ell.$$
