Home Competitions AUSTRIAN-POLISH 1992 Problem 8
AUSTRIAN-POLISH 1992 Problem 8
Let n ≥ 3 n\ge 3 n ≥ 3 a 1 , . . . , a n a_1,..., a_n a 1 , ... , a n − a 1 − a 2 + a 3 + . . . a n a 1 = a 1 − a 2 − a 3 + a 4 + . . . a n a 2 = . . . = a 1 + . . . + a n − 2 − a n − 1 − a n a n − 1 = − a 1 + a 2 + . . . + a n − 1 − a n a n \frac{-a_1-a_2+a_3+...a_n}{a_1}=\frac{a_1-a_2-a_3+a_4+...a_n}{a_2}=...=\frac{a_1+...+a_{n-2}-a_{n-1}-a_n}{a_{n-1}}=\frac{-a_1+a_2+...+a_{n-1}-a_n}{a_{n}} a 1 − a 1 − a 2 + a 3 + ... a n = a 2 a 1 − a 2 − a 3 + a 4 + ... a n = ... = a n − 1 a 1 + ... + a n − 2 − a n − 1 − a n = a n − a 1 + a 2 + ... + a n − 1 − a n a 2 + a 3 + . . . a n a 1 ⋅ a 1 + a 3 + a 4 + . . . a n a 2 ⋅ . . . ⋅ + a 1 + a 2 + . . . + a n − 1 a n \frac{a_2+a_3+...a_n}{a_1}\cdot \frac{a_1+a_3+a_4+...a_n}{a_2}\cdot ...\cdot \frac{+a_1+a_2+...+a_{n-1}}{a_{n}} a 1 a 2 + a 3 + ... a n ⋅ a 2 a 1 + a 3 + a 4 + ... a n ⋅ ... ⋅ a n + a 1 + a 2 + ... + a n − 1
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