Áp dụng tính chất dãy tỉ số bằng nhau ta có:
$\frac{a}{a+b}=\frac{b}{b+c}=\frac{c}{c+a}=\frac{a+b+c}{a+b+b+c+c+a}=\frac{a+b+c}{2\left( a+b+c \right)}=\frac{1}{2}$
$\Rightarrow \left\{ \begin{align}
& \frac{a}{a+b}=\frac{1}{2} \\
& \frac{b}{b+c}=\frac{1}{2} \\
& \frac{c}{c+a}=\frac{1}{2} \\
\end{align} \right.$$\Rightarrow \left\{ \begin{align}
& 2a=a+b \\
& 2b=b+c \\
& 2c=c+a \\
\end{align} \right.$$\Rightarrow a=b=c$
Khi đó $S=\frac{a+b}{2c}+\frac{b+c}{3a}+\frac{c+a}{4b}$
$S=\frac{a+a}{2a}+\frac{a+a}{3a}+\frac{a+a}{4a}=1+\frac{2}{3}+\frac{1}{2}=\frac{13}{6}$