The condition gives

$$bc-b-c=a(1-b-c),\qquad bc-a=(b+c)(1-a).$$

Also, $b+c\ne1$: otherwise the condition would give $bc=1$, contradicting $bc\le(b+c)^2/4=1/4$. Since $a\ne1$, both factors in the denominator are nonzero. Thus

$$\frac{(bc-b-c)(1-a)}{(bc-a)(1-b-c)}=\frac a{b+c}.$$

The other two fractions simplify in the same way. It remains to prove

$$\frac a{b+c}+\frac b{c+a}+\frac c{a+b}\ge\frac{a+b+c}{2}.$$