1 x 2 x n 1 x n 2 x n 1 h ? n n ữ a ta có đi u ph

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1 + x 2 > x n 1 x n > 2 x n 1 . H ơ n n a, Ta có đi u ph i ch ng minh. 2 < x n y n = x n x n 1 < 3 . Bài toán 10(Romania 1999 [10]) Cho A = { a 1 , a 2 , · · ·} ⊂ N là t p th a mãn v i m i dãy con phân bi t B , C A , x = x . Ch ng minh r ng: x B x C 1 1 1 a 1 + a 2 + · · · + a n < 2 . L i gi i Tr c ướ h t ta ch ng minh b đ sau: B đ : Cho ế dãy x 1 , y 1 , x 2 , y 2 , · · ·, x n , y n các s th c d ng th a mãn: ươ (i), x 1 y 1 < x 2 y 2 < · · · < x n y n ; (ii), x 1 + x 2 + · · · + x k y 1 + y 2 + · · · + y k v i m i k = 1 , 2 , · · ·, n . Khi đó ta có : 1 1 1 1 1 1 x 1 + x 2 + · · · + x n y 1 + y 2 + · · · + y n . Th t v y ta có: Đ t i = 1 x i y i k , i = x i y i v i m i 1 i n . Ta có th gi s ả ử 1 > 2 > · · · > n > 0 và i i 0 v i m i 1 k n . Chú ý r ng: n . 1 1 n Σ 2 n 1 n 1 1 3 n 1 ƒ = k =
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k = 1 n n 1 y k x k = k k = n i + ( k k + 1 )( 1 + 2 + · · · + k ) ≥ 0 , i = 1 k = 1
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V y b đ đ c ch ng minh. ượ Quay l i bài toán, không m t tính t ng quát ta gi s ả ử a 1 < a 2 < · · · < a n , và đ t y i = 2 i 1 v i m i i . Rõ ràng: a 1 y 1 < a 2 y 2 < · · · < a n y n . V i k b t kỳ, t ng 2 k 1 đ c t o ra b ng cách ch n ít nh t m t trong dãy r i r c ượ a 1 , a 2 , · · ·, a k k . Do v y l n nh t trong đó i = 1 a i , nh nh t là 2 k 1. Do v y v i k = 1 , 2 , · · ·, n ta có: a 1 + a 2 + · · · + a k 2 k 1 = y 1 + y 2 + · · · + y k . Áp d ng b đ , ta có: 1 1 1 1 1 1 1 a 1 + a 2 + · · · + a n < y 1 + y 2 + · · · + y n = 2 2 n 1 < 2 . Đây là đi u c n ch ng minh. Bài toán 11(Bulgaria 1999 [8]) . Ch ng minh r ng v i s nguyên b t kỳ n , n 3, t n t i n s nguyên d ng ươ a 1 , a 2 , · · ·, a n trong c p s c ng, n s nguyên d ng ươ b 1 , b 2 , · · ·, b n trong c p s nhân, sao cho: b 1 < a 1 < b 2 < a 2 < · · · < b n < a n . Đ a ra ví d v 2 dãy này v i m i dãy là ý nh t 5 s h ng. ư L i gi i Ta tìm dãy mà b n = a n 1 + 1 và b n 1 = a n 2 + 1. Vi t ế d = a n 1 a n 2 . Khi đó v i 2 i , j n 1 ta có b b b b = d , sao cho b = b n 1 ( b b ) > i + 1 i a n 1 + ( n j ) d = a j 1 . n n 1 j n i i = j i + 1 Và n u ta ch c ch n ế b a , khi đó b b j 1 b b a j 1 d a 1 < 1 j = 1 + ( i = 1 i + 1 +
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i ) ≤ 1 + ( − ) = j
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v i m i j , nên chu i b t đ ng th c đã đ c th a mãn. ượ Cho b 1 , b 2 , · · ·, b n b ng k n 1 , k n 2 ( k + 1 ), · · ·, k 0 ( k + 1 ) n 1 , đó k là m t giá tr đã đ c xác đ nh sau đó. ượ Ta cũng đ t a n 1 = b n 1 và a n 2 = b n 1 1, và đ nh nghĩa a i khác theo đó. Khi đó d = a n a n 1 = b n b n
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