Balkan Shortlist, 2012

We will prove that there exists a polynomial $P(x)$ which satisfies the given condition and has $2012$ distinct real roots.
First we note that the given inequality is equivalent to
$$(P(a) + P(b) + P(c))((P(a) - P(b))^2 + (P(b) - P(c))^2 + (P(c) - P(a))^2) \geq 0,$$
so it is enough to find a polynomial $P$ such that $P(a)+P(b)+P(c) \geq 0$ whenever $a+b+c = 0$.
For positive numbers $M$ and $\varepsilon$ let
$$P_{M,\varepsilon}(x) = (x - M)(x - M - \varepsilon)\cdots(x - M - 2011\varepsilon).$$
$P_{M,\varepsilon}$ is positive and decreasing on $(-\infty, M)$, and $P_{M,\varepsilon}$ is positive and increasing on $(M + 2011\varepsilon, \infty)$. We have $P_{M,\varepsilon}(x) \geq M^{2012}$ for $x \leq 0$ and
$$|P_{M,\varepsilon}(x)| = |(x - M) \cdot (x - M - \varepsilon) \cdots (x - M - 2011\varepsilon)| \leq (2011\varepsilon)^{2012}$$
for $x \in [M, M + 2011\varepsilon]$. Therefore $P_{M,\varepsilon}(x) \geq -(2011\varepsilon)^{2012}$ for $x \geq 0$.
Let $a, b$ and $c$ be real numbers with $a + b + c = 0$. Without loss of generality assume that $a \leq 0$. From the previous inequalities we have
$$P_{M,\varepsilon}(a) + P_{M,\varepsilon}(b) + P_{M,\varepsilon}(c) \geq M^{2012} + 2(-(2011\varepsilon)^{2012}).$$
Since the right hand side of the last inequality is positive for $M = 2$ and $\varepsilon = 1/2011$, we can take $P(x)$ to be $P_{2,1/2011}(x)$.

Note that $P(a)^3 + P(b)^3 + P(c)^3 \geq 3P(a)P(b)P(c)$ follows from the AM-GM inequality if $P(a), P(b), P(c)$ are all nonnegative.
We will again work with $P(x) = P_{M,\varepsilon}(x)$ and we may again assume that $a \leq 0$.
If only one of $P(b)$ and $P(c)$ is negative then we have
$$P(a)^3 + P(b)^3 + P(c)^3 \geq M^{3\cdot 2012} - (2011\varepsilon)^{3\cdot 2012} \geq 0 \geq 3P(a)P(b)P(c)$$
if $M \geq 2011\varepsilon$.
On the other hand, if both $P(b)$ and $P(c)$ are negative, then
$$P(a)^3 + P(b)^3 + P(c)^3 \geq P(a)^3 - 2(2011\varepsilon)^{3\cdot 2012} \geq 3(2011\varepsilon)^{2\cdot 2012}P(a) \geq 3P(a)P(b)P(c)$$
if $M \geq 2^{1/2011}2011\varepsilon$ as $u^3 - 2v^3 - 3v^2u = (u - 2v)(u + v)^2 \geq 0$ for $u \geq 2v$.
We conclude again that $P(x) = P_{2,1/2011}(x)$ works.