Serbian selection contest for the BMO 2025 - P4

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Serbian selection contest for the BMO 2025 - P4

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#1 h Apr 7, 2025, 10:15 PM

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Let $a_1, a_2, \ldots, a_8$ be real numbers. Prove that
$\sum_{i=1}^{8} \left( a_i^2 + a_i a_{i+2} \right) \geq \sum_{i=1}^{8} \left( a_i a_{i+1} + a_i a_{i+3} \right),$ where the indices are taken modulo 8, i.e., $a_9 = a_1$ , $a_{10} = a_2$ , and $a_{11} = a_3$ . In which cases does equality hold?

Proposed by Vukašin Pantelić and Andrija Živadinović

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#3 h Apr 10, 2025, 9:55 AM

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$\textbf{Property a): }x^2+y^2\ge2xy\Longleftrightarrow (x-y)^2\ge0,\;\forall\;x,y\in\mathbb{R}$ , with equality for $x=y$ .
$\textbf{Property b): }\sum_{i=1}^{8}a_i a_{i+5}=\sum_{i=1}^{8}a_i a_{i-3}=\sum_{i=1}^{8}a_i a_{i+3}$ .

$\textbf{Relation 1): }$
$2\cdot \sum_{i=1}^{8} \left( a_i^2 + a_i a_{i+2} \right)=\sum_{i=1}^{8} \left( a_i^2+2a_ia_{i+2}+a_{i+2}^2 \right)=$
$=\sum_{i=1}^{8} \left( a_i+a_{i+2} \right)^2$ .

$\textbf{Relation 2): }$
$2\cdot\sum_{i=1}^{8} \left( a_i+a_{i+2} \right)^2=\sum_{i=1}^{8} \left(\left( a_i+a_{i+2} \right)^2+\left( a_{i+1}+a_{i+3} \right)^2\right)\ge\;\textbf{(Property a)}$
$\ge 2\sum_{i=1}^{8}\left( a_i+a_{i+2} \right)\cdot\left( a_{i+1}+a_{i+3} \right)=2\sum_{i=1}^{8}\left( a_ia_{i+1}+a_ia_{i+3}+a_{i+1}a_{i+2}+a_{i+2}a_{i+3} \right)=$
$=6\cdot\sum_{i=1}^{8}a_i a_{i+1}+2\cdot\sum_{i=1}^{8}a_{i}a_{i+3}$ .

$\textbf{Relation 3): }$
$2\cdot\sum_{i=1}^{8} \left( a_i+a_{i+2} \right)^2=\sum_{i=1}^{8} \left(\left( a_i+a_{i+2} \right)^2+\left( a_{i+3}+a_{i+5} \right)^2\right)\ge\;\textbf{(Property a)}$
$\ge 2\sum_{i=1}^{8}\left( a_i+a_{i+2} \right)\cdot\left( a_{i+3}+a_{i+5} \right)=$
$=2\sum_{i=1}^{8}\left( a_ia_{i+3}+a_ia_{i+5}+a_{i+2}a_{i+3}+a_{i+2}a_{i+5} \right)=\;\textbf{(Property b)}$
$=2\cdot\sum_{i=1}^{8}a_{i}a_{i+1}+6\cdot\sum_{i=1}^{8}a_i a_{i+3}$ .

Using the relations $\textbf{1), 2), 3)}$ results:
$8\cdot \sum_{i=1}^{8} \left( a_i^2 + a_i a_{i+2} \right)=2\cdot\sum_{i=1}^{8} \left( a_i+a_{i+2} \right)^2+2\cdot\sum_{i=1}^{8} \left( a_i+a_{i+2} \right)^2\ge$
$\ge \left(6\cdot\sum_{i=1}^{8}a_i a_{i+1}+2\cdot\sum_{i=1}^{8}a_{i}a_{i+3}\right)+\left(2\cdot\sum_{i=1}^{8}a_{i}a_{i+1}+6\cdot\sum_{i=1}^{8}a_i a_{i+3}\right)=$
$=8\cdot\sum_{i=1}^{8}\left(a_{i}a_{i+1}+a_i a_{i+3}\right)\Longleftrightarrow$
$\Longleftrightarrow \sum_{i=1}^{8} \left( a_i^2 + a_i a_{i+2} \right) \geq \sum_{i=1}^{8} \left( a_i a_{i+1} + a_i a_{i+3} \right)$ .

$\textbf{The equality occurs for:}$
$a_i+a_{i+2}=a_{i+1}+a_{i+3}$ and $a_i+a_{i+2}=a_{i+3}+a_{i+5},\;\forall\;i\in\{1,2,...,7,8\}\Longleftrightarrow$
$\Longleftrightarrow a_{i+5}=a_{i+1}$ and $a_i+a_{i+2}=a_{i+1}+a_{i+3}=S,\;\forall\;i\in\{1,2,...,7,8\}\Longleftrightarrow$
$\Longleftrightarrow (a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8)=(a,b,S-a,S-b,a,b,S-a,S-b)$ , with $a,b,S\in\mathbb{R}$ .

This post has been edited 1 time. Last edited by WallyWalrus, Apr 10, 2025, 9:57 AM

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