2002 Pan African MO Problems/Problem 6

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Problem

If $a_1 \geq a_2 \geq \cdots \geq a_n \geq 0$ and $a_1+a_2+\cdots+a_n=1$, then prove: \[a_1^2+3a_2^2+5a_3^2+ \cdots +(2n-1)a_n^2 \leq 1\]

Solution

Note that $1 = (a_1 + a_2 + \cdots + a_n)^2 = \sum_{i=1}^{n} (a_i^2) + 2a_1a_2 + 2a_1a_3 + \cdots 2a_{n-1}a_n$. Additionally, if $i \le j$, then $a_i \ge a_j$ and $2a_i a_j \ge 2a_j^2$.


For a given value $j$, there are $j-1$ terms in the form $2a_ia_j$. Thus, \begin{align*} (\sum_{i=1}^n a_i)^2 &\ge \sum_{i=1}^{n} (a_i^2) + 2a_2^2 + 4a_3^2 + \cdots + (2n-2)a_n^2 \\ 1 &\ge a_1^2 + 3a_2^2 + 5a_3^2 + \cdots + (2n-1)a_n^2. \end{align*}

See Also

2002 Pan African MO (Problems)
Preceded by
Problem 5
1 2 3 4 5 6 Followed by
Last Problem
All Pan African MO Problems and Solutions
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