Library / Twentieth IMAR Mathematical Competition /1 of 3

Twentieth IMAR Mathematical Competition

Algebra Difficulty 4.7 AIME Prove it Romania

Fix integers $n \ge 2$ and $1 \le m \le n-1$ . Let $a_0$ , $a_1$ , $\dots$ , $a_n$ be non-negative real numbers satisfying $a_0 + a_1 + \dots + a_n = 1$ . Prove that, if $\sum_{k=0}^n a_k x^k < x^m$ for some $0 < x < 1$ , then $\sum_{k=0}^{m-1} (m-k)a_k < \sum_{k=m+1}^n (k-m)a_k$ .

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Solution

As $a_0 + a_1 + \dots + a_n = 1$ , the required inequality is equivalent to $\sum_{k=0}^n k a_k > m$ . To prove this inequality, note that the exponential $t \mapsto x^t$ , $t \in \mathbb{R}$ , is convex and apply Jensen's inequality to write $x^{\sum_{k=0}^n k a_k} \le \sum_{k=0}^n a_k x^k < x^m$ . As $0 < x < 1$ , the desired inequality follows by comparing the exponents of $x$ at both ends.

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