If $0\le a,b,c\le 1$, then $\frac a{1+bc}+\frac b{1+ac}+\frac c{1+ab}\le 2$

Note that $a,b,c \in [0,1] \implies abc \le ab,bc,ca$.

Thus, $\displaystyle \frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab} \le \frac{a}{1+abc}+\frac{b}{1+abc}+\frac{c}{1+abc} = \frac{a+b+c}{1+abc}$

Since, $\displaystyle a,b,c \in [0,1] \implies a(1-b)(1-c) \ge 0 \implies \sum\limits_{cyc} a(1-b)(1-c) \ge 0 \\ \implies a+b+c + 3abc \ge 2(ab+bc+ac)$

$$\begin{align} \implies 2(abc+1) &\ge 2abc + 2 + 2(ab+bc+ac) - 3abc - (a+b+c) \\ &= a+b+c + abc + 2(1-a)(1-b)(1-c) \\ &\ge a+b+c \end{align}$$

Thus, $\displaystyle \frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab} \le \frac{a+b+c}{1+abc} \le 2$

Let $f(a,b,c)=\sum\limits_{cyc}\frac{a}{1+bc}$.

Thus, $f$ is a convex function of $a$, of $b$ and of $c$.

Id est, $$\max_{\{a,b,c\}\subset[0,1]}f=\max_{\{a,b,c\}\subset\{0,1\}}f=f(1,1,0)=2.$$