Đặt $x= \frac{b+c-a}{2} ; y= \frac{c+a-b}{2} ; z= \frac{a+b-c}{2} (x, y,z >0)$ $\Rightarrow$ $x+y+z= \frac{a+b+c}{2}$ =1
$\Rightarrow a=y+z ; b=z+x ;c =x+y$
$\Rightarrow S=\frac{y+z}{2x} +\frac{4(z+x)}{2y} +\frac{9(x+y)}{2z} $
= $\frac{1}{2}(\frac{y}{x} + \frac{4x}{y} + \frac{z}{x} + \frac{9x}{z}+ \frac{4z}{y}+ \frac{9y}{z})$
$\geq\frac{1}{2} (4+6+12) =11$
Dấu $"=" \Leftrightarrow$ $a=\frac{5}{6}; b=\frac{2}{3}; c=\frac{1}{2}$.