$a+b>c\implies 2(a+b)>a+b+c\implies \frac{2c}{2(a+b)} < \frac{2c}{a+b+c}$ Similary $\frac{2b}{2(a+c)} < \frac{2b}{a+b+c}$ and $\frac{2a}{2(b+c)} < \frac{2a}{a+b+c}\implies \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} < \frac{2a}{a+b+c)}+ \frac{2b}{a+b+c}+\frac{2c}{a+b+c} = 2$

nesbitts inequality