Solutions - Georg Mohr-Konkurrencen

√ √It has a root 1 √2 < α < 1, because 79 + 1 7 + 19 = > 1 and 7 39 + 1 9 < 1.We will prove that a n M · n α for some M > 0 — since nαwill benarbitrarily small for large enough n, the claim follows from this immediately.We choose M so that the inequality a n M · n α holds for 1 n 8;since for n 9 we have 1 < [7n/9] < n and 1 [n/9] < n, it follows byinduction that[ 7n] α [ n] αa n = a [7n/9] + a [n/9] M · + M · 99( 7n) α ( n) ( α (7 ) α ( ) 1 α M · + M · = M · nα · + = M · n999 9) α .14. Let the numbers be x 1 x 2 . . . x 2n−1 x 2n . We will show that thechoice s 1 = x 1 + x 3 + x 5 + · · · + x 2n−1 and s 2 = x 2 + x 4 + · · · + x 2n solvesthe problem. Indeed, the inequality s 1s 2 1 is obvious and we haves 1s 2= x 1 + x 3 + x 5 + . . . + x 2n−1x 2 + x 4 + x 6 + . . . + x 2n= (x 3 + x 5 + . . . + x 2n−1 ) + x 1(x 2 + x 4 + . . . + x 2n−2 ) + x 2n (x 3 + x 5 + . . . + x 2n−1 ) + 1(x 2 + x 4 + . . . + x 2n−2 ) + 2 (x 2 + x 4 + . . . + x 2n−2 ) + 1(x 2 + x 4 + . . . + x 2n−2 ) + 2 =1= 1 −(x 2 + x 4 + . . . + x 2n−2 ) + 2 1 − 1(n − 1) + 2 = nn + 1 .15. Let i 2. We are given the inequalities(i − 1) · a 2 i−1 i · a i a i−2 (5)andi · a 2 i (i + 1) · a i+1 a i−1 . (6)Multiplying both sides of (6) by x 2 , we obtaini · x 2 · a 2 i (i + 1) · x 2 · a i+1 a i−1 . (7)10