Ukrainian National Mathematical Olympiad, 4th Round

We can rewrite the given equality as follows:
$$4abc - 4(ab + bc + ca) + 4(a+b+c) - 4 = -(ab + bc + ca) + 2(a+b+c) - 3,$$
$$4(a-1)(b-1)(c-1) = -(a-1)(b-1) - (b-1)(c-1) - (c-1)(a-1),$$
$$4 = \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c}.$$
The function $f(x) = \frac{1}{1-x}$ is convex on the interval $x < 1$ because $f''(x) = \frac{2}{(1-x)^3} \ge 0$. So, by the Jensen's inequality,
$$4 = f(a) + f(b) + f(c) \ge 3f\left(\frac{a+b+c}{3}\right) = \frac{9}{3-(a+b+c)}.$$
The last inequality implies the required one.