Ukrainian National Mathematical Olympiad, 3rd Round

It is enough to prove the inequality for $a + b + c = 2$.
Let us denote $x = ab + bc + ac$, then
$$a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ac) = 4 - 2(ab + bc + ac) = 2(2 - x).$$
From the obvious inequality $x(2 - x) \leq 1$, we get
$$\begin{aligned} 2 + 2abc &\geq 2 \geq 2x(2 - x) = (ab + bc + ac)(a^2 + b^2 + c^2) = \\ &= ab(a^2 + b^2) + cb(c^2 + b^2) + ac(a^2 + c^2) + abc(a + b + c), \end{aligned}$$
Using the equality $2abc = abc(a + b + c)$, we arrive at the conclusion.

Equality occurs, for example, $a = b = 1$, $c = 0$.