Arbitrage bet on $3$-way horse race 
There are $3$ horses in this week's race. When we say "the stated odds against a horse winning are $r$-to-$s$", we mean if you wager $1$ dollar on the horse, you will lose $1$ dollar if the horse loses and win $r/s$ dollars if the horse wins. The stated odds against the horse $A$ winning are $2$-to-$1$. The stated odds against $B$ winning are $3$-to-$1$. If the odds against horse $C$ are $4$-to-$1$, how can you get to ensure you will win money?


I did some guesswork: Intuitively since one of the three has to win, we should pile the most money onto horse $A$ since it has the lowest payout if it wins. $2$-$1$-$1$ didn't work, so then I tried $3$-$2$-$2$ and it magically worked.
But I am wondering how one could deduct this more systematically. Here's my attempt. Let $X$ be the total amount of money I bet, $X_A$, $X_B$, $X_C$ the amounts I bet on each horse. We want:
$$ 2X_A > X_A + X_B + X_C, \qquad 3 X_B > X_A + X_B + X_C, \qquad 4X_C > X_A + X_B + X_C.$$
However, I am not sure what to do from here. Can anyone please help?
 A: Letting $a,b,c$ be the amounts of money you make on each bet, the correct inequalities are
$$
2a>b+c\qquad 3b>a+c\qquad4c>a+b
$$
since a winning bet only needs to recoup the losses from the other bets.
A way to simplify this is to note that all the inequalities are scale invariant; if $a,b,c$ satisfy the above, then so will $ra,rb$ and $rc$, for any $r>0$. To eliminate this redundancy, let
$$
x=b/a,\qquad y=c/a
$$
and notice the inequalities become
$$
2>x+y,\qquad 3x>1+y,\qquad 4y>1+x.
$$
It is simple to graphically solve these inequalities, and determine the set of $(x,y)$ which works is the interior of a certain triangle: https://www.desmos.com/calculator/e1vjvdpwh6. For example, it is pretty clear that $(2/3,2/3)$ is inside that triangle, which leads to the solution $(a,b,c)=(3,2,2)$.
A: Let $a$, $b$ and $c$ be the amounts bet on horses $A$, $B$ and $C$, respectively. You pay $a+b+c$, and you win $3a$, $4b$ or $5c$, depending on which horse wins. I say $3, 4, 5$ instead of $2,3,4$ to have the same data as you, because I want to separate phase 1 (what you bet) from phase 2 (what you win).
You want to win $\$100$ in all options. So, you have to bet $\$100/3 = \$33.33$ on $a$, plus $\$100/4 = \$25$ on $b$ and $\$100/5=\$20$ on $c$. Total, you bet $33.33+25+20=78.33$, and you are sure to win $100$. The profit is $\$21.67$, no matter the winner.
In real bets, you will never have this configuration, you need to pay more than $\$100$ if you want to be sure to win $\$100$.
