$\underline{\textbf{solution 1}}$ By AM-GM, we have $$\frac{x}{y} + \frac{y}{x}\ge 2\cdot\sqrt{\frac{x}{y}}\cdot\sqrt{\frac{y}{x}}=2$$ Therefore, $$\Big(\frac{a+b}{b+c}+\frac{b+c}{a+b}\Big)+\Big(\frac{b+c}{c+a}+\frac{c+a}{b+c}\Big)+\Big(\frac{c+a}{a+b}+\frac{a+b}{c+a}\Big) \ge 2+2+2=6$$ The left side can be re-arranged as \begin{align} &\Big(\frac{a+b}{b+c}+\frac{c+a}{b+c}\Big)+\Big(\frac{b+c}{c+a}+\frac{a+b}{c+a}\Big)+\Big(\frac{c+a}{a+b}+\frac{b+c}{a+b}\Big)\\ &=1+\frac{2a}{b+c}+1+\frac{2b}{c+a}+1+\frac{2c}{a+b}\\ &=3 + 2\Big(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\Big) \end{align} $$\therefore\quad 3 + 2\Big(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\Big)\ge 6$$ which is equivalent to $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2} $$ $\underline{\textbf{Solution 2}}$ By AH-HM, we have \begin{align} \frac{(a+b)+(b+c)+(c+a)}{3}&\ge\frac{3}{\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}} \end{align} $$\therefore\quad {\Big((a+b)+(b+c)+(c+a)\Big)}{\Big(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\Big)}\ge9$$ Simplifying this relation will lead to the desired result.