Let $f(x)=x^n+a_1x^{n-1}+\ldots+a_n~(n\ge3)$ be a polynomial with real coefficients and $n$ real roots, such that $\frac{a_{n-1}}{a_n}>n+1$. Prove that if $a_{n-2}=0$, then at least one root of $f(x)$ lies in the open interval $\left(-\frac12,\frac1{n+1}\right)$.
Problem
Source: Bulgaria 1987 P1
Tags: algebra, Polynomials, Sequences, polynomial
orangecurtis
09.07.2022 20:41
Let $f(x) = (x-r_1)(x-r_2)...(x-r_n).$ Substitute $b_i = \frac{1}{r_i}$ $\forall$ $1 \le i \le n.$ In summary by the problem conditions, we need $\sum_{1 \le i \le n}b_i < -n-1$ and $(\sum_{1 \le i\le n}b_i)^{2} = \sum_{1 \le i \le n}b_i^{2} $ and it's desired to show that not all $b_i \in (-2, n+1).$
For the sake of contradiction, assume that for all $1 \le i \le n, b_i \in (-2, n+1).$ If $n$ was an odd positive integer $2m-1,$ then $b_1 + b_2 + ... + b_{2m-1} < -2m \Rightarrow$ there must be at least $m+1$ $b_i$'s that are negative. Additionally, $(b_1 + b_2 + ... + b_{2m-1})^{2} > 4m^{2}.$ Now find an upper bound on $b_1^2 + b_2^2 + ... + b_{2m-1}^2.$ By Greedy Algorithm, all negative $b_i$ terms should equal -2 and assume WLOG that $b_{m-k}, b_{m-k+1}, ... , b_{2m-1}$ all equal -2 so that there are $(m+k)$ -2's and $(m-k-1)$ $b_i$'s are positive for some $m-1 \ge k, $ where $k$ is a positive integer. Now using the fact that $b_1 + b_2 + ... + b_{m-k-1} < 2k:$
\begin{align*}
&b_1^2 + b_2^2 + ... + b_{2m-1}^2 \le b_1^2 + b_2^2 + ... + b_{m-k-1}^2 + 4(m+k) < 4k^2 + 4(m+k)\\
&\Longleftrightarrow b_1^2 + b_2^2 + ... + b_{2m-1}^2 < 4(m-1)^2 + 4m + 4(m-1)\\
&\Longleftrightarrow b_1^2 + b_2^2 + ... + b_{2m-1}^2 < 4m^2
\end{align*}Thus $(\sum_{1 \le i \le 2m-1}b_i)^2 \neq \sum_{1 \le i \le 2m-1}b_i^2$ and we have a contradiction, thus our original assumption must be wrong.
The situation for even $n$ is pretty much exactly the same except you plug in $n = 2m$ and work with that. The same contradiction then arises and in summary we must have that not all $b_i \in (-2, n+1) \,$ i.e., there are some $r_i \in (-\frac{1}{2}, \frac{1}{n+1}),$ as desired.