Inequalities and rearrangements Some children are arranged in two rows, so that each child in the front row is taller than the child behind him in the back row. The children are now rearranged in increasing order in each row. Show that in the new arrangement  each child in the front row is taller than the child behind him.
Reformulation: consider two $n$-uples of (positive) numbers $(a_1, \ldots, a_n)$ and $(b_1, \ldots, b_n)$ such that $a_i \ge b_i$ for all $1\le i \le n$. If $(a'_1, \ldots, a'_n)$ and $(b'_1, \ldots, b'_n)$ are the increasing rearrangements of $(a_1, \ldots, a_n)$ respectively $(b_1, \ldots, b_n)$ then  $a'_i \ge b'_i$ for all $1\le i\le n$. 
I was told that the statement is obvious. Is it? It is fairly easy to see that $a'_1 \ge b'_1$, and also that $a'_n \ge b'_n$, but it does not seem obvious for other values of $i$. 

Look at it inductively.  The shortest kid in the front ($a'_1$) is clearly taller than the shortest kid in back ($b'_1$) . Why?  Well, $a'_1$ was taller than some kid in back and that kid had to be at least as tall as $b'_1$.
Now, suppose (inductively) that we are done for the first $i-1$ kids.  Look at the original configuration.  If $a'_1, ...a'_{i-1}$ originally were in front of the list $b'_1, ...b'_{i-1}$ (in any order) then clearly the kid $a'_i$ was originally in front of someone no shorter than $b'_i$.  Alternatively, suppose that at least one of $a'_1, ...a'_{i-1}$ was in front of some b other than one of $b'_1, ...b'_{i-1}$.  But that would mean that there were more than $i-1$ kids in the back row shorter than $a'_{i-1}$ which means that $b'_i < a'_{i-1} < a'_i$ as desired.
