2 (n 1). f (i 1, i 2,..., i k )

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1 Math Time problem proposal #1 Darij Grinberg version 5 December 2010 Problem. Let x 1, x 2,..., x n be real numbers such that x 1 x 2...x n 1 and such that x i < 1 for every i {1, 2,..., n}. Prove that 1 x i 1 x j n 2 n 1. 1 n Soluon. First, for the sake of brevity 1, we introduce a notaon: If i 1, i 2,..., i k are some free variables, if f is a funcon of k variables, and if A 1, A 2,..., A l are some asserons containing some of the variables i 1, i 2,..., i k, then i 1, i 2,..., i k A 1, A 2,..., A l f i 1, i 2,..., i k will mean the sum of the values f i 1, i 2,..., i k over all good k-tuples i 1, i 2,..., i k ; hereby, a k-tuple i 1, i 2,..., i k is called good if it sasfies i u {1, 2,..., n} for every u {1, 2,..., k} and also sasfies all the asserons A 1, A 2,..., A l. Using this notaon, we can write many typical sums in a simpler form: For instance, f i i f i, and f i, j f. 1 i n 1 n Note that we won t use the notaon v ku f k for f u f u 1... f v in this soluon, since it theorecally could be misunderstood because different meaning in our above notaon. With the notaon defined above, we have 1 n 1 x i 1 x j 1 x i 1 x j. Hence, the inequality in queson, 1 x i 1 x j n 2 n 1, rewrites as 1 n 1 x i 1 x j n 2 n 1. Mulplicaon by 2 n 1 2 transforms this inequality into n x i 1 x j 1 In as far as brevity is possible for this soluon... n n 1. v f k has a ku 1

2 Since 2 1 x i 1 x j this inequality rewrites as n x i 1 x j 1 x i 1 x j j<i 1 x i 1 x j x j x i 1 x j 1 x i at this step we have interchanged i with j in the second sum 1 x i 1 x j 1 x i 1 x j, j<i 1 x i 1 x j 1 x i 1 x j n n 1. 1 Thus, in order to solve the problem, it remains to prove this inequality 1. Set 1 x i for every i {1, 2,..., n}. Then, > 0 for every i {1, 2,..., n} since x i < 1 and thus 1 x i > 0, so that 1 x i > 0. Besides, 1 x i yields x i 1. Finally, k tk k 1 xk n k xk n x 1 x 2... x n n

3 Consequently, n x i 1 x j n 1 x i n 1 x j 1 x i 1 x j n 1 1 n 1 1 k t k n 1 m t m n 1 since 2 yields n 1 k tk n 1 n 1 n 1 n 1 k t k and n 1 k t k k m t m m 1 since n 1 k and n 1 k t k m m k t k m k t k 1 Now, k t k k t k m m k tk m k m t k 1 n n 1 k t k m m tm. 3 because n n 1 is the number of all pairs with i and j being elements of the m t m 3

4 set {1, 2,..., n} and sasfying i j. Besides, k t k m m m here we have interchanged i with j and renamed k by m in the first sum m t m i 1 j m i j 1 m j m t m i m 1 j j m i m 1 j 2 t m 2 n 1 t m n i j 1 n 1 j i i j 1 n 1 Hence, 3 becomes n }{{} nn 1 n n 1 j n 1 j n x i 1 x j k t k m }{{} 0 k m t k. here we have renamed m by j in the second sum i k t k Therefore, the inequality that we have to prove, namely 1, rewrites as n n 1 This is obviously equivalent to k t k k t k m n n 1. m m 0. 4 Thus, we only have to prove 4 in order to complete the soluon of the problem because 4 is equivalent to 1, and proving 1 solves the problem. 4

5 Now, k t k, k, m m,,, k i, m j, k, m,,, k, m,,, ki mj }{{} 0, since we have ki or mj and thus t k 0 or t m 0 because every quadruple, k, m sasfies exactly one of the two asserons k i m j and k i m j, k, m,,, k i, m j, k, m,,, k i, m j, k m, k, m,,, k i, m j, k m 0, k, m,,, k i, m j, k, m,,, k i, m j, km, k,, k i t }{{ j } t k t k, since km t k t k. Thus, in order to prove 4, is enough to prove the two inequalies, k, m,,, k i, m j, k m, k,, k i 0; 5 t k t k 0. 6 Proving 5 is rather easy: First, the six condions i j, k j, m i, k i, m j, k m altogether connected by a logical and are equivalent to the condion that the four numbers, k, m are pairwisely disnct, i. e. that the set {, k, m} 5

6 has exactly 4 elements, i. e. that {, k, m} 4. Hence,, k, m,,, k i, m j, k m , k, m {i,j,k,m} 4, k, m {i,j,k,m} 4, k, m {i,j,k,m} 4, k, m {i,j,k,m} 4 t k t m t m, k, m {i,j,k,m} 4, k, m {i,j,k,m} 4, k, m {i,j,k,m} 4, k, m {i,j,k,m} 4, k, m {i,j,k,m} 4 t k t k t k t m t k t m at this step, we have renamed some variables: in the second sum, we interchanged i with k; in the third sum, we interchanged j with m; in the fourth sum, we interchanged both i with k and j with m, k, m {i,j,k,m} 4, k, m {i,j,k,m} 4, k, m {i,j,k,m} 4, k, m {i,j,k,m} 4 tk t k t k t k t m t m t k t m t k t m tk t k t k t m t k t m t k t m t k t m t k t m 1 4, k, m {i,j,k,m} 4 t k 2 2 t k t m 0 because squares are nonnegave; thus, 5 is proven. In order to prove 6, we proceed similarly: First, the three condions i j, k j, k i altogether connected by a logical and are equivalent to the condion that the three numbers, k are pairwisely disnct, i. e. that the set {, k} has exactly 6

7 3 elements, i. e. that {, k} 3. Hence,, k,, k i t k t k, k {i,j,k} 3, k {i,j,k} 3, k {i,j,k} 3, k {i,j,k} 3, k {i,j,k} 3 t k t k t k t k t k t k, k {i,j,k} 3, k {i,j,k} 3 t k t k t k t k, k {i,j,k} 3, k {i,j,k} 3 at this step, we have renamed some variables: in the second sum, we renamed, k by j, k, i, respecvely; in the third sum, we renamed, k by k,, respecvely tk t k t k t k t k t k t k t k t k t k t k t k 0, t k t k t k t k because the Schur inequality yields t k t k t k t k t k 0 for any triple, k. Thus, the inequality 6 is proven. As both inequalies 5 and 6 are verified now, the inequality 4 follows, and thus the problem is solved. Remark. The above problem is a generalizaon of the Schur inequality which states that a a b a c b b c b a c c a c b 0 for any three nonnegave reals a, b, c to n variables in fact, if you set n 3 in the problem, you get the Schur inequality for the numbers 1 x 1, 1 x 2, 1 x 3. In the parcular case when the reals x 1, x 2,..., x n are nonnegave, the problem can be solved much more easily - a soluon for this case was given in [1] proof of Theorem 4.1. References [1] Darij Grinberg, An inequality involving 2n numbers. 7

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