IMO 1996 SL A2

Origin: IRE | Category: Algebra

Problem

Let $a_1 \geq a_2 \geq \cdots \geq a_n$ be real numbers such that

$a_1^k + a_2^k + \cdots + a_n^k \geq 0$

for all integers $k > 0$. Let $p = \max{|a_1|, \ldots, |a_n|}$. Prove that $p = a_1$ and that

$(x - a_1)(x - a_2) \cdots (x - a_n) \leq x^n - a_1 x^{n-1} + \cdots + (-1)^n a_n$

for all $x > a_1$.

Solution

Clearly $a_1 > 0$, and if $p \neq a_1$, we must have $a_n < 0$, $|a_n| > |a_1|$, and $p = -a_n$. But then for sufficiently large odd $k$,

$(-a_n)^k = |a_n|^k > (n-1)|a_1|^k,$

so that

$a_1^k + \cdots + a_n^k \leq (n-1)|a_1|^k - |a_n|^k < 0,$

a contradiction. Hence $p = a_1$.

Now let $x > a_1$. From $a_1 + \cdots + a_n \geq 0$ we deduce

$\sum_{j=2}^n (x - a_j) \leq (n-1) \frac{x + a_1}{n-1} = x + a_1,$

so by the AM–GM inequality,

(x - a_2) \cdots (x - a_n) \leq \left( \frac{x + a_1}{n-1} \right)^{n-1} \leq x^{n-1} + x^{n-2} a_1 + \cdots + a_1^{n-1} \tag{1}

The last inequality holds because

$\binom{n-1}{r} \leq (n-1)^r \quad \text{for all } r \geq 0.$

Multiplying (1) by $(x - a_1)$ yields the desired inequality.