Conc. interj. "Gheorghe Lazar" Sibiu 2010 probl. 2

[Andi Brojbeanu]

Determinati numerele reale \( x_i, i=\overline{1,n}, n\ge 2 \), stiind ca \( \sum_{i=1}^n {x_i}=a \) si \( \sum_{1\le i<j\le n}{x_i x_j} =\frac{n-1}{2n} a^2, a\in \mathb{R} \).
Dumitru Acu, Sibiu

[Andi Brojbeanu]

\( \sum_{i=1}^n {x_i^2}=(\sum_{i=1}^n {x_i})^2-2\sum_{1\le i<j\le n} {x_ix_j}=a^2-2\cdot \frac{n-1}{2n}a^2=a^2(1-\frac{n-1}{n})=\frac{a^2}{n} \).
Dar, cu inegalitatea CBS avem: \( \sum_{i=1}^n {x_i^2}\ge \frac{(\sum_{i=1}^n {x_i})^2}{n}=\frac{a^2}{n} \).
Deci, avem egalitate. Atunci \( x_1=x_2=....=x_n=\frac{a}{n} \).