IMC 1998 · Day 1 · P5

Observe that both sides of (2) are identically equal to zero if $n = 1$. Suppose that $n > 1$. Let $x_1, \dots, x_n$ be the zeros of $P$. Clearly (2) is true when $x = x_i$, $i \in \{1, \dots, n\}$, and equality is possible only if $P'(x_i) = 0$, i.e., if $x_i$ is a multiple zero of $P$. Now suppose that $x$ is not a zero of $P$. Using the identities $$\frac{P'(x)}{P(x)} = \sum_{i=1}^{n} \frac{1}{x - x_i}, \qquad \frac{P''(x)}{P(x)} = \sum_{1 \le i < j \le n} \frac{2}{(x - x_i)(x - x_j)},$$ we find $$(n-1) \left( \frac{P'(x)}{P(x)} \right)^2 - n\, \frac{P''(x)}{P(x)} = \sum_{i=1}^{n} \frac{n-1}{(x - x_i)^2} - \sum_{1 \le i < j \le n} \frac{2}{(x - x_i)(x - x_j)}.$$ But this last expression is simply $$\sum_{1 \le i < j \le n} \left( \frac{1}{x - x_i} - \frac{1}{x - x_j} \right)^2,$$ and therefore is positive. The inequality is proved. In order that (2) holds with equality sign for every real $x$ it is necessary that $x_1 = x_2 = \dots = x_n$. A direct verification shows that indeed, if $P(x) = c (x - x_1)^n$, then (2) becomes an identity.