$$\begin{array}{l} (a+b+c)\left(\frac{1}{bc+1}+\frac{1}{ca+1}+\frac{1}{ab+1}\right)= \\ \frac{a}{bc+1}+\frac{b}{ca+1}+\frac{c}{ab+1}+\frac{b+c}{bc+1}+\frac{c+a}{ca+1}+\frac{a+b}{ab+1} \leqslant \\ \frac{a}{bc+1}+\frac{b}{ca+1}+\frac{c}{ab+1}+1+1+1 \leqslant \\ \frac{a}{bc+1}+\frac{b}{ca+b}+\frac{c}{ab+c}+3= \\ \frac{a}{bc+1}-\frac{ca}{ca+b}-\frac{ab}{ab+c}+5= \\ a\left(\frac{1}{bc+1}-\frac{c}{ca+b}-\frac{b}{ab+c}\right)+5 \leqslant \\ a\left(1-\frac{c}{c+b}-\frac{b}{b+c}\right)+5=5 \end{array}$$

Equality holds if and only if $a=0, b=c=1$ and its cyclic permutations, thus the proposition is proved!