This should rather be a comment, posting it as an answer is to give a visual clue that there is something in the vein of a proof.

Elaborating on Timothy Chow's insightful comment it suffices to consider a single pair of people, $A.$ and $B.$ of which $A.$ has the two coins $a_0$ and $a_1,\ a_0\le a_1$ of smaller value; likewise $B.$'s coins are $b_0$ and $b_1,\ b_0\le b_1;\quad a_0+a_1\lt b_0+b_1,$.

under the assumption that $a_1\gt b_1$ we have:
$a_0+a_1=b_0+b_1-\Delta,\quad 0\lt\Delta\lt b_0+b_1$
$a_0\ =\ b_0+b_1-\Delta-a_1\ <\ b_0+b_1-\Delta-b_1\ =\ b_0-\Delta\quad\implies\quad a_0+\Delta\lt b_0$

which proves that the assumption implies that the smaller coin of the smaller amount must be less than the smaller coin of the larger value.

This should rather be a comment, posting it as an answer is to give a visual clue that there is something in the vein of a proof.

Elaborating on Timothy Chow's insightful comment it suffices to consider a single pair of people, $A.$ and $B.$ of which $A.$ has the two coins $a_0$ and $a_1,\ a_0\le a_1$ of smaller value; likewise $B.$'s coins are $b_0$ and $b_1,\ b_0\le b_1;\quad a_0+a_1\lt b_0+b_1$.

Under the assumption that $a_1\gt b_1$ we have:
\begin{align*} & a_0+a_1=b_0+b_1-\Delta,\quad 0\lt\Delta\lt b_0+b_1 \\ & a_0\ =\ b_0+b_1-\Delta-a_1\ <\ b_0+b_1-\Delta-b_1\ =\ b_0-\Delta\quad\implies\quad a_0+\Delta\lt b_0 \end{align*} which proves that the assumption implies that the smaller coin of the smaller amount must be less than the smaller coin of the larger value.

This should rather be a comment, posting it as an answer is to give a visual clue that that there is something in the vein of a proof.

Elaborating on Timothy Chow's insightful comment it suffices to consider a single pair of people, $A.$ and $B.$ of which $A.$ has the two coins $a_0$ and $a_1,\ a_0\le a_1$ of smaller value; likewise $B.$'s coins are $b_0$ and $b_1,\ b_0\le b_1;\quad a_0+a_1\lt b_0+b_1,$.

under the assumption that $a_1\gt b_1$ we have:
$a_0+a_1=b_0+b_1-\Delta,\quad 0\lt\Delta\lt b_0+b_1$
$a_0\ =\ b_0+b_1-\Delta-a_1\ <\ b_0+b_1-\Delta-b_1\ =\ b_0-\Delta\quad\implies\quad a_0+\Delta\lt b_0$

which proves that the assumption implies that the smaller coin of the smaller amount must be less than the smaller coin of the larger value.

This should rather be a comment, posting it as an answer is to give a visual clue that there is something in the vein of a proof.

Elaborating on Timothy Chow's insightful comment it suffices to consider a single pair of people, $A.$ and $B.$ of which $A.$ has the two coins $a_0$ and $a_1,\ a_0\le a_1$ of smaller value; likewise $B.$'s coins are $b_0$ and $b_1,\ b_0\le b_1;\quad a_0+a_1\lt b_0+b_1,$.

under the assumption that $a_1\gt b_1$ we have:
$a_0+a_1=b_0+b_1-\Delta,\quad 0\lt\Delta\lt b_0+b_1$
$a_0\ =\ b_0+b_1-\Delta-a_1\ <\ b_0+b_1-\Delta-b_1\ =\ b_0-\Delta\quad\implies\quad a_0+\Delta\lt b_0$

which proves that the assumption implies that the smaller coin of the smaller amount must be less than the smaller coin of the larger value.