Baltic Way 1999 - Georg Mohr-Konkurrencen

If n = m 3 is a solution, then m satisfies 1000m m 3 < 1000(m + 1).From the first inequality, we get m 2 1000, or m 32. By the secondinequality, we then havem 2 < 1000 · m + 1m33 1000 1000 · = 1000 +32 32 1032 ,or m 32. Hence, m = 32 and n = m 3 = 32768 is the only solution.3. Answer: n = 3 and n = 4.For n = 3 we havea 1 a 2 + a 2 a 3 + a 3 a 1 = (a 1 + a 2 + a 3 ) 2 − (a 2 1 + a 3 2 + a 2 3)2 (a 1 + a 2 + a 3 ) 22= 0 .For n = 4, applying the AM-GM inequality we havea 1 a 2 + a 2 a 3 + a 3 a 4 + a 4 a 1 = (a 1 + a 3 )(a 2 + a 4 ) (a 1 + a 2 + a 3 + a 4 ) 24= 0 .For n 5 take a 1 = −1, a 2 = −2, a 3 = a 4 = · · · = a n−2 = 0, a n−1 = 2,a n = 1. This givesa 1 a 2 + a 2 a 3 + . . . + a n−1 a n + a n a 1 = 2 + 2 − 1 = 3 > 0 .( 14. Answer: the maximum value is f √2 ,1)√ = √ 1 . 2 2We shall make use of the inequality x 2 + y 2 2xy . If x x yx 2 + y 2 y2xy = 12x ,yx 2 + y 2 , thenimplying x 1 √2, and the equality holds if and only if x = y = 1 √2.4