Note that

$$1-\frac{a-b c}{a+b c}=\frac{2 b c}{1-b-c+b c}=\frac{2 b c}{(1-b)(1-c)}$$

The inequality is equivalent to

$$\frac{2 b c}{(1-b)(1-c)}+\frac{2 c a}{(1-c)(1-a)}+\frac{2 a b}{(1-a)(1-b)} \geq \frac{3}{2} .$$

Manipulation yields the equivalent

$$4(b c+c a+a b-3 a b c) \geq 3(b c+c a+a b+1-a-b-c-a b c) .$$

This simplifies to $a b+b c+c a \geq 9 a b c$ or

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 9 \text {. }$$

This is a consequence of the harmonic-arithmetic means inequality.