Balkan 2012 shortlist

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 $$(P(a)+P(b)+P(c))((P(a)-P(b){)}^2+(P(b)-P(c){)}^2+(P(c)-P(a){)}^2)\ge 0,$$ so it is enough to find a polynomial $P$ such that $P(a)+P(b)+P(c)\ge 0$ whenever $a+b+c=0$. For positive numbers $M$ and $\epsilon$ let $$P_{M,\epsilon }(x)=(x-M)(x-M-\epsilon )\cdots (x-M-2011\epsilon ).$$ $P_{M,\epsilon }$ is positive and decreasing on $(-\infty ,M)$, and $P_{M,\epsilon }$ is positive and increasing on $(M+2011\epsilon ,\infty )$. We have $P_{M,\epsilon }(x)\ge M^{2012}$ for $x\le 0$ and $$|P_{M,\epsilon }(x)|=|(x-M)\cdot (x-M-\epsilon )\cdots (x-M-2011\epsilon )|\le (2011\epsilon {)}^{2012}$$ for $x\in [M,M+2011\epsilon ]$. Therefore $P_{M,\epsilon }(x)\ge -(2011\epsilon {)}^{2012}$ for $x\ge 0$. Let $a,b$ and $c$ be real numbers with $a+b+c=0$. Without loss of generality assume that $a\le 0$. From the previous inequalities we have $$P_{M,\epsilon }(a)+P_{M,\epsilon }(b)+P_{M,\epsilon }(c)\ge M^{2012}+2(-(2011\epsilon {)}^{2012}).$$ Since the right hand side of the last inequality is positive for $M=2$ and $\epsilon =1/2011$, we can take $P(x)$ to be $P_{2,1/2011}(x)$.

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Alternative solution.

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