Let $P(x)$ be a polynomial of degree $2012$ with real coefficients satisfying the condition \[P(a)^3 + P(b)^3 + P(c)^3 \geq 3P(a)P(b)P(c),\] for all real numbers $a,b,c$ such that $a+b+c=0$. Is it possible for $P(x)$ to have exactly $2012$ distinct real roots?
Problem
Source: Serbia Additional TST 2012, Problem 1
Tags: algebra, polynomial, function, inequalities, absolute value, algebra proposed
19.05.2012 22:12
Just take $P(x)=(x-2012)(x-2013)(x-2014)\cdots)(x-4023).$
05.02.2013 15:21
and how do you know that the condition is satisfied?
05.02.2013 21:16
Assume $a \ge b \ge c$ If none of the numbers is in $[2012,4023]$ it is trivial as it is all positif. If one is, its function value is a lot of smaller in absolute value than $P(c)$ and so $P(c)^3 > 3P(c)^2 P(3017.5)$ which will solve it .
26.01.2015 00:11
But how do you know the condition is true for all $a,b,c$? Besides, SCP wrote: If none of the numbers is in $[2012,4023]$ it is trivial as it is all positif. What is positive?
26.01.2015 00:40
P(a),P(b) en p(c)
27.01.2015 03:13
If they are positive, how would that prove the problem?
27.01.2015 20:56
MathPanda1 wrote: If they are positive, how would that prove the problem? Then the inequality becomes obvious because $x^3+y^3+z^3 \ge 3xyz$ trivially holds for $x,y,z \ge 0$ by AM-GM.
27.01.2015 23:22
I knew that (how could I be so stupid?). So, I was just wondering how when it isn't all positive, SCP's argument proves the problem (or am I just being dumb again).
12.10.2020 17:11
SCP wrote: If one is, its function value is a lot of smaller in absolute value than $P(c)$ and so $P(c)^3 > 3P(c)^2 P(3017.5)$ which will solve it . I don't understand this. Could you please explain more ? Also, does anyone have another solutions ?
07.01.2025 05:47
Does this work?