A Mediterranean polynomial has only real roots and it is of the form \[ P(x) = x^{10}-20x^9+135x^8+a_7x^7+a_6x^6+a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0 \] with real coefficients $a_0\ldots,a_7$. Determine the largest real number that occurs as a root of some Mediterranean polynomial. (Proposed by Gerhard Woeginger, Austria)
Problem
Source: MMO 2011 Problem #1
Tags: algebra, polynomial, Vieta, inequalities, algebra unsolved
hyperbolictangent
12.09.2011 05:12
Let the real roots of the polynomial be \[a \le b \le c \dots \le j\] Then by Vieta, \[a + b + c + \dots j = 20\] \[a^2 + b^2 + c^2 + \dots j^2 = (a + b + c + \dots j)^2 - 2(ab + ac + \dots ij) = 20^2 - 2(135) = 130\]
Thus \[a + b + c + \dots i = 20 - j\] \[a^2 + b^2 + \dots i^2 = 130 - j^2\]
By Cauchy-Schwarz, \[(a^2 + b^2 + c^2 + \dots i^2)(1^2 + 1^2 + \dots 1^2) \ge (a + b + c + \dots i)^2\]\[(130 - j^2)(9) \ge (20 - j)^2\] \[10(j - 11)(j + 7) \le 0\]
Thus the maximum real value of $j$ is $11$.
It is easy to see that \[(x - 11)(x - 1)^9 = x^{10} - 20x^9 + 135x^8 + \dots\] does in fact these conditions, and so the maximum possible value of a real root is $11$.
woe
16.04.2013 14:21
Consider a Mediterranean polynomial that has a real root $\alpha$ and nine other real roots $x_1,\ldots,x_9$. Let $s=\sum_{i=1}^9x_i$, and $t=\sum_{i=1}^9x_i^2$, and $u=\sum_{1\le i<j\le 9}x_ix_j$. Viete's theorem implies $s=20-\alpha$ and $u=135-s\alpha=135-20\alpha+\alpha^2$. Furthermore $t=s^2-2u=130-\alpha^2$. Observe that \[ 0 ~\le~ \sum_{1\le i<j\le 9} (x_i-x_j)^2 ~=~ 8t-2u ~=~ 10(\alpha+7)(11-\alpha).\] This inequality implies $\alpha\le11$. Since $\alpha=11$ and $x_1=x_2=\cdots=x_9=1$ yield a Mediterranean polynomial with root 11, the answer to the problem is 11.
LMat
11.09.2015 15:19
This problem is basically asking "Given $10$ values with mean $2$ and variance $9$ what is the greatest possible deviation from the mean?" This is a well known problem. https://en.wikipedia.org/wiki/Samuelson%27s_inequality