IMC 2013 · Day 1 · P4

Let $$p(X) = \prod_{i=1}^{n} (X - x_i) = X^n - A X^{n-1} + \frac{A^2 - B}{2} X^{n-2} - \frac{A^3 - 3AB + 2C}{6} X^{n-3} + \dots.$$ The $(n-3)$th derivative of $p$ has three nonnegative real roots $0 \le u \le v \le w$. Hence, $$\frac{6}{n!}\, p^{(n-3)}(X) = X^3 - \frac{3A}{n} X^2 + \frac{3(A^2 - B)}{n(n-1)} X - \frac{A^3 - 3AB + 2C}{n(n-1)(n-2)} = (X - u)(X - v)(X - w),$$ so $$u + v + w = \frac{3A}{n}, \quad uv + vw + wu = \frac{3(A^2 - B)}{n(n-1)} \quad \text{and} \quad uvw = \frac{A^3 - 3AB + 2C}{n(n-1)(n-2)}.$$ From these we can see that $$\begin{align*} \frac{n^2 (n-1)^2 (n-2)}{9} &\bigl( (n+1) A^2 B + (n-2) B^2 - A^4 - (2n-2) AC \bigr) = \dots = \\ &= u^2 v^2 + v^2 w^2 + w^2 u^2 - uvw(u + v + w) = uv(u-w)(v-w) + vw(v-u)(w-u) + wu(w-v)(u-v) \ge \\ &\ge 0 + uw(v-u)(w-v) + wu(w-v)(u-v) = 0. \end{align*}$$