Let $a_i, b_j \in \mathbb{N}$ for all $ 1 \leq i, j \leq n$. Suppose $a_i \leq b_i$ for any $i$. Define $$\alpha _ 1 = \min(a_1, \ldots, a_n), \ \beta_1 = \min (b_1, \ldots, b_n)$$ and
$$\alpha_i = \min \{(a_1, a_2, \ldots, a_n) - \alpha_1 - \alpha_2 - \cdots - \alpha_{i-1}\}, \\ \beta_i = \min \{(b_1, b_2, \ldots, b_n) - \beta_1 - \beta_2 - \cdots - \beta_{i-1}\}$$
for all $2 \leq i \leq n,$ where $(a_1, a_2, \ldots, a_n) - a_i = (a_1, a_2, \ldots, a_{i-1}, a_{i+1}, \ldots, a_n)\ $ for any $1 \leq i \leq n.$ Then $$\alpha_i \leq \beta_i$$ for any $i$.
Roughly, it is the comparison of the $i$ minimum of $(a_i), (b_j).$
I try to prove by induction. The basis step is clear : $\alpha_1 = \min (a_i) \leq b_j$ for any $j$. So $\alpha_1 \leq \beta_1$
Inductive step : Assume $\alpha_i \leq \beta_i$ for any $i \leq k$. I will show that $\alpha_{k+1} \leq \beta_{k+1}.$
Try both direct proof and contradiction proof (cannot reach contradiction), but I cannot finish the argument.
Any help please ?
Here's something slightly informal:
So w.l.o.g. take $\alpha_1=a_1.\quad$ (otherwise one could relabel)
If $\beta_1=b_1$, then by applying the same argument as you did with $n$ replaced by $n-1$, (with renaming or relabelling if you wish), on the sets $\{a_2,\ldots, a_n\}$ and $\{b_2,\ldots, b_n\}$.
It follows that $\alpha_i\leq\beta_i$.
Now if $\beta_1\neq b_1$, say w.l.o.g. $\beta_1=b_2$, then notice that:
$a_1=\alpha_1\leq a_2\leq b_2 =\beta_1\leq b_1$,
thus we have in the new sets $\{a_2,\ldots, a_n\}$ and $\{b_1,b_3,\ldots,b_n\}$ that for $i\geq 3$, $a_i\leq b_i$, and also $a_2\leq b_1$.
We can then simply relabel these sets as $\{a'_1,\ldots, a'_{n-1}\}$ and $\{b'_1,b'_2,\ldots,b'_{n-1}\}$ in the obvious way, and again apply the same argument.