# A Polynomial-Inequality problem from Vietnam National Olympiad 2021

Abstract

I have just completed 2 days of our National Olympiad. This year's problems are not difficult, yet new and strange to many of the students. Mentioned below will be problem 5 out of 7, the first problem of Day 2.

To anyone who wonders why it is VMO 2021, we always do national olympiad for a year before the previous year ends, so that the result will be published in early 2021

Problem

Given that $$P(x)=a_{21}x^{21}+a_{20}x^{20}+a_{19}x^{19}+ \dotsb + a_1x+a_0$$ is a polynomial with real coefficients such that $$a_i \in [1011,2021]$$ for all $$i \in \{ 0,1,2,\dotsb,21\}$$ and $$c \in \mathbb{R}$$ such that $$| a_{k+2}-a_{k} | \le c$$ for all $$k \in \{ 0,1,2, \dotsb,19 \}$$ Given that $$P(x)$$ has at least one integer root.

1. Prove that $$P(x)$$ has exactly one integer root
2. Prove the following inequality: $$\displaystyle\sum_{k=0}^{10} |a_{2k+1}-a_{2k}|^2 \le 440c^2$$

My works

1. For part 1, it is easy to see that since $$a_i$$ are all positive, then if $$P(x)$$ has a root, that root must be negative. From the condition $$a_i \in [1011,2021]$$ it is also easy to see that $$\frac{a_i}{a_j} <2$$ for all $$0 \le i,j \le 21$$, thus if $$x$$ is an integer root of $$P(x)$$ then $$x \ge (-2)$$. Thus $$x = (-1)$$ is the only integer root of $$P(x)$$.

2. For part 2, some of my teammates say that it could be solved using Jensen inequality, but none of us succeeded. I noticed that $$\displaystyle\sum_{k=0}^{10} a_{2k+1} =\displaystyle\sum_{k=0}^{10} a_{2k}$$, but how do I finish the problem?

Any help is appreciated.

• Are you able to find a (close to) equality case, say for $c = 1$? I suspect that we have a much tighter bound of $60c$. Dec 26, 2020 at 15:09
• Well I doubt that 440 could not be the best coefficient to use but I still think that the power of $c$ in the inequality must be at least 2, which means it's not that tight as $60c$ (unless you can give the solution to it). Moreover, just to remind, we can not compare $60c$ to $440c^2$ as $c$ runs through the real axis. Dec 26, 2020 at 15:15
• Right, after realizing that $c$ is real, I believe I have a counter example. Can you check? Dec 26, 2020 at 15:17
• Sorry I posted an edit to the inequality. My memory was bad Dec 26, 2020 at 15:19
• With your edit, 440 is the best coefficient, and I have stated an equality case.IIRC there is a pretty nice approach for this problem (without the polynomial skin). Dec 26, 2020 at 15:20

(This is before the edit.)

The inequality is not true.

Take $$c = 0.1$$ (Any $$0 < c < \frac{3}{22}$$ will work.)

Take $$a_{2k} = 1011 + kc$$, $$a_{2k+1} = 1011 + (10-k) c$$.
This satisfies the conditions as 1) $$P(-1) = 0$$ is an integer root, 2) The difference between terms is exactly $$c$$.

However, $$\sum |a_{2k+1} - a_{2k}| = 10c + 8c + \ldots + 0c + 2c + \ldots + 10c = 60c > 440 c^2$$.

My suspicion is that $$\sum |a_{2k+1} - a_{2k}| \leq 60c$$, though I don't know how to prove it. The equality case is as above.

Claim: If

• $$\sum_{i=0}^{21} (-1)^i a_i = 0$$
• $$|a_{k+2} - a_k | \leq |c|$$

Then $$\sum_{k=0}^{10} |a_{k+1} - a_k |^2 \leq 440 c^2$$.

Equality case is when $$a_{2k} = a + kc$$, $$a_{2k+1} = a + (10-k) c$$.

• yes, that was my bad, i posted the wrong equality. It must be $\sum |a_{2k+1}-a_{2k}|^2$. Can you please resolve it? Dec 26, 2020 at 15:19