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Team Selection Test for the 54th IMO

Algebra Difficulty 8.2 Shortlist Prove it Bulgaria

Let $n \ge 2$ be a positive integer and $a_1 < a_2 < \dots < a_{2n}$ be real numbers. If $S = \sum_{i=1}^{2n} a_i$ , $A_1 = \sum_{i,j,i<j} a_{2i}a_{2j}$ and $A_2 = \sum_{i,j,i<j} a_{2i-1}a_{2j-1}$ , prove the inequality
$(n-1)S^2 > 4n(A_1 + A_2).$

Solution

First, we shall prove the following
Lemma. If $P(x) = b_0x^n + b_1x^{n-1} + b_2x^{n-2} + \dots + b_{n-1}x + b_n$ has $n$ real distinct roots then $(n-1)b_1^2 - 2nb_0b_2 > 0$ .

Proof. First differentiate $n-2$ times the function $f(x)$ . As a result we have quadratic function having two real distinct roots and therefore its discriminant is positive.

Consider the polynomial $P(x) = (x-a_1)(x-a_3)\dots(x-a_{2n-1})+(x-a_2)(x-a_1)\dots(x-a_{2n})$ . It is straightforward to verify that it satisfies the condition of the lemma and the corresponding inequality is exactly the desired inequality
$(n-1)S^2 > 4n(A_1 + A_2).$

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