IMC 2003 · Day 1 · P6

The polynomial $f$ is a product of linear and quadratic factors, $f(z) = \prod_i (k_i z + l_i) \cdot \prod_j (p_j z^2 + q_j z + r_j)$, with $k_i, l_i, p_j, q_j, r_j \in \mathbb{R}$. Since all roots are in the left half-plane, for each $i$, $k_i$ and $l_i$ are of the same sign, and for each $j$, $p_j, q_j, r_j$ are of the same sign, too. Hence, multiplying $f$ by $-1$ if necessary, the roots of $f$ don't change and $f$ becomes the polynomial with all positive coefficients.

For the simplicity, we extend the sequence of coefficients by $a_{n+1} = a_{n+2} = \dots = 0$ and $a_{-1} = a_{-2} = \dots = 0$ and prove the same statement for $-1 \le k \le n-2$ by induction.

For $n \le 2$ the statement is obvious: $a_{k+1}$ and $a_{k+2}$ are positive and at least one of $a_{k-1}$ and $a_{k+3}$ is 0; hence,

$a_{k+1} a_{k+2} > a_k a_{k+3} = 0$.

Now assume that $n \ge 3$ and the statement is true for all smaller values of $n$. Take a divisor of $f(z)$ which has the form $z^2 + pz + q$ where $p$ and $q$ are positive real numbers. (Such a divisor can be obtained from a conjugate pair of roots or two real roots.) Then we can write $$\tag{1} f(z) = (z^2 + pz + q)(b_{n-2} z^{n-2} + \dots + b_1 z + b_0) = (z^2 + pz + q) g(x).$$ The roots polynomial $g(z)$ are in the left half-plane, so we have $b_{k+1} b_{k+2} < b_k b_{k+3}$

for all $-1 \le k \le n-4$. Defining $b_{n-1} = b_n = \dots = 0$ and $b_{-1} = b_{-2} = \dots = 0$ as well, we also have $b_{k+1} b_{k+2} \le b_k b_{k+3}$

for all integer $k$.

Now we prove $a_{k+1} a_{k+2} > a_k a_{k+3}$. If $k = -1$ or $k = n-2$ then this is obvious since $a_{k+1} a_{k+2}$ is positive and $a_k a_{k+3} = 0$. Thus, assume $0 \le k \le n-3$. By an easy computation, $$\begin{align*} a_{k+1} &a_{k+2} - a_k a_{k+3} = \\ &= (q b_{k+1} + p b_k + b_{k-1})(q b_{k+2} + p b_{k+1} + b_k) - (q b_k + p b_{k-1} + b_{k-2})(q b_{k+3} + p b_{k+2} + b_{k+1}) = \\ &= (b_{k-1} b_k - b_{k-2} b_{k+1}) + p (b_k^2 - b_{k-2} b_{k+2}) + q (b_{k-1} b_{k+2} - b_{k-2} b_{k+3}) + \\ &\quad + p^2 (b_k b_{k+1} - b_{k-1} b_{k+2}) + q^2 (b_{k+1} b_{k+2} - b_k b_{k+3}) + pq (b_{k+1}^2 - b_{k-1} b_{k+3}). \end{align*}$$ We prove that all the six terms are non-negative and at least one is positive. Term $p^2 (b_k b_{k+1} - b_{k-1} b_{k+2})$ is positive since $0 \le k \le n-3$. Also terms $b_{k-1} b_k - b_{k-2} b_{k+1}$ and $q^2 (b_{k+1} b_{k+2} - b_k b_{k+3})$ are non-negative by the induction hypothesis.

To check the sign of $p (b_k^2 - b_{k-2} b_{k+2})$ consider $$b_{k-1} (b_k^2 - b_{k-2} b_{k+2}) = b_{k-2} (b_k b_{k+1} - b_{k-1} b_{k+2}) + b_k (b_{k-1} b_k - b_{k-2} b_{k+1}) \ge 0.$$ If $b_{k-1} > 0$ we can divide by it to obtain $b_k^2 - b_{k-2} b_{k+2} \ge 0$. Otherwise, if $b_{k-1} = 0$, either $b_{k-2} = 0$ or $b_{k+2} = 0$ and thus $b_k^2 - b_{k-2} b_{k+2} = b_k^2 \ge 0$. Therefore, $p (b_k^2 - b_{k-2} b_{k+2}) \ge 0$ for all $k$. Similarly, $pq (b_{k+1}^2 - b_{k-1} b_{k+3}) \ge 0$.

The sign of $q (b_{k-1} b_{k+2} - b_{k-2} b_{k+3})$ can be checked in a similar way. Consider $$b_{k+1} (b_{k-1} b_{k+2} - b_{k-2} b_{k+3}) = b_{k-1} (b_{k+1} b_{k+2} - b_k b_{k+3}) + b_{k+3} (b_{k-1} b_k - b_{k-2} b_{k+1}) \ge 0.$$ If $b_{k+1} > 0$, we can divide by it. Otherwise either $b_{k-2} = 0$ or $b_{k+3} = 0$. In all cases, we obtain $b_{k-1} b_{k+2} - b_{k-2} b_{k+3} \ge 0$.

Now the signs of all terms are checked and the proof is complete.