51st 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 $f^{\prime \prime }(x)=\frac{2}{(1-x{)}^3}\ge 0$. 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.