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
SCP
19.05.2012 22:12
Just take $P(x)=(x-2012)(x-2013)(x-2014)\cdots)(x-4023).$
anonymouslonely
05.02.2013 15:21
and how do you know that the condition is satisfied?
SCP
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 .
MathPanda1
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?
SCP
26.01.2015 00:40
P(a),P(b) en p(c)
MathPanda1
27.01.2015 03:13
If they are positive, how would that prove the problem?
Tintarn
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.
MathPanda1
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).
KrysTalk
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 ?
AshAuktober
07.01.2025 05:47
Does this work?
Indeed, consider any monic polynomial
$$Q(x) := \prod_{i = 1}^{2012} (x-a_{i}),$$where $a_1 < a_2 < \dots < a_{2012}$ are the roots of $Q$. Note that $Q$ obtains a global minimum $g < 0$.
Now consider a real value $r_0 > a_{2012}$ such that $Q'(r) > 0 \forall r \ge r_0$, and $Q(r_0) \ge 2|g|$.
We define $Q(x) \equiv P(x-r_0)$, and claim that $Q$ satisfies the given condition (in particular, we prove that $a + b + c = 0 \implies P(a) + P(b) + P(c) \ge 0$.
We consider cases, letting WLOG $a \le b \le c$.
Case 1: $a < 0, b, c, \ge 0$.
Note that $P(a) \ge g,$ and $P(b), P(c) \ge P(0) \ge 2|g|$, so $P(a) + P(b) + P(c) \ge 2|g|+g = |g| > 0$, as desired.
Case 2: $a, b < 0, c \ge 0$.
Then we have $P(a), P(b) \ge g,$ and $P(c) \ge 2|g|$, so $P(a) + P(b) + P(c) \ge 2|g| + 2g = 0$ as desired.
Case 3: $a, b, c \ge 0$.
Then $a, b, c = 0$, for which the result is trivial as $P(0) > 0$.
Therefore we have indeed proven that the answer is $\boxed{\text{Yes}}$.
The first lesson this problem teaches is to never assume the answer is always no, a presumption that I definitely might have had in the back of my mind while solving this problem. Because if you're adamant on proving the answer is no, then you're gonna struggle a lot, because you're trying to prove something that isn't true.
Now, onto the actual solution.
I kind of tried to get intuition about the problem by seeing which polys don't work. And, well, the first poly that comes to mind is one with a negative leading coefficient. Okay, so let's assume this leading coefficient is positive, then. So the graph of P increases a lot as x becomes very negative or very positive. This is a good start, as we want $P(a) + P(b) + P(c) \ge 0$ or $P(a) = P(b) = P(c)$, and it seems likely the former holds. This gives a confirmation the answer should be yes.
The only issue, though, is that the roots might be very close to zero, in which case everything is uncertain if you take all $a, b, c$ close to 0. And then the key idea is to notice you can actually just avoid this from happening entirely (!!) by shifting the graph of $P$ to the left as much as required, so that the stuff to the right of zero is so huge that the negative stuff in between the roots doesn't even matter! (This is what was up with the whole $r_0$ business in my solution.) And from here, it's just converting words into algebra, which isn't too bad.