Nesbitt's inequality

Substitution. Let $x=b+c$, $y=c+a$, $z=a+b$. Then $a=(y+z-x)/2$, $b=(z+x-y)/2$, $c=(x+y-z)/2$, and
$$\frac{a}{b+c}=\frac{y+z-x}{2x}=\frac12\left(\frac{y}{x}+\frac{z}{x}-1\right).$$
Summing cyclically:
$$S=\frac12\left[\left(\frac{y}{x}+\frac{z}{x}+\frac{z}{y}+\frac{x}{y}+\frac{x}{z}+\frac{y}{z}\right)-3\right].$$

AM-GM. By AM-GM, $\frac{y}{x}+\frac{x}{y}\ge2$ and similarly for the other pairs. So the sum in brackets is $\ge6$.

Conclusion. $S\ge\frac12(6-3)=\frac32$. Equality iff $x=y=z$, i.e. $a=b=c$.