Int. Journal of Math. Analysis, Vol. 6, 2012, no. 38, 1871 - 1875
Reverse Shapiro Type Inequality Tanfer Tanriverdi Department of Mathematics, Harran University, Sanliurfa, 63300 Turkey
[email protected] Abstract We propose the following inequality which has not been been studied yet. Tn (x) =
n xk+1 + xk+2 k=1
xk
≥ 2n,
where all xk > 0, and xn+k = xk for k ∈ N. Examples show that Tn (x) is indeed true for all n with an analytic proof.
Mathematics Subject Classification: 26D05, 26D15, 26D20 Keywords: Cyclic inequality, Shapiro’s inequality
1
Introduction
Shapiro [1] conjectured that E(x) =
n k=1
n xk ≥ , xk+1 + xk+2 2
(1)
where xk ≥ 0, xk+1 + xk+2 > 0 and xn+k = xk for k ∈ N. To understand the importance of new inequality we briefly go over the other side of it. Studies on (1) have been based on counterexamples and analytic proofs have been given for small n so far. Several authors [3] proved that (1) is true for n = 3, 4, 5, 6. Diananda [2] proved that (1) is true for n ≤ 6 different from the previous ones. Mordell [4] proved that (1) is true for n = 7. Dojokovic [5] proved that (1) is true for n = 8. Nowosad [6] proved that (1) is true for n = 10. Bushell and Craven [7] also proved that (1) is true for n = 10. Godunova and Levin [8] verified that (1) is true for n = 12 partly analytically and partly numerically. Recently, Bushell and Mcleod [9] proved analytically that (1) is true for n = 12.
1872
T. Tanriverdi
Analytic proof for n = 23 is still open. Troesch [10, 11] numerically proved that (1) is true for n = 13, 23. For more sophisticated analysis and a brief history on conjecture see [6, 9, 12]. In this paper we analyze the following inequality which I called reverse Shapiro type inequality (RSTI). Tn (x) =
n xk+1 + xk+2 k=1
xk
≥ 2n,
(2)
where all xk > 0, and xn+k = xk for k ∈ N.
2
Main Results
Set F (k) =
xk+1 + xk+2 , xk
xk > 0 where k = 1, 2, . . . , n.
(3)
Then n
F (k) =
k=1
(x2 + x3 ) . . . (xn + xn+1 )(xn+1 + xn+2 ) . x1 x2 . . . xn−1 xn
(4)
Using the arithmetic mean and geometric mean inequality we reformulate RSTI in terms of given data as follows. n
F (k) ≥ n{
k=1
(x2 + x3 ) . . . (xn + xn+1 )(xn+1 + xn+2 ) 1/n } , x1 x2 x3 . . . xn
(5)
where xn+1 = x1 and xn+2 = x2 . So we formally proved the following theorem. Theorem 2.1. Assume all xk are hold as in (2). Then n
F (k) ≥ n{
k=1
(x2 + x3 ) . . . (xn + xn+1 )(xn+1 + xn+2 ) 1/n } . x1 x2 x3 . . . xn
Corollary 2.2. If all xk > 0 are equal then one gets equality in (2). Theorem 2.3. If all xk > 0 are equal or different from each other then (2) holds. √ Proof. Applying (xk + xk+1 ) ≥ 2 xk xk+1 where k = 1, 2, . . . , n to the numerator of (4) with xn+1 = x1 and xn+2 = x2 , one obtains
So
n
(x1 + x2 )(x2 + x3 ) . . . (xn + x1 ) ≥ 2n x1 x2 x3 . . . xn−1 xn .
k=1 F (k)
≥ 2n . From (5) we have the result as announced.
1873
Reverse Shapiro type inequality
We now give and analyze some other properties of (2) with Tn (x) = T (x). Lemma 2.4. Tn (x) is a homogeneous function of degree 0. Proof. Proof is trivial. Homogeneous functions have wide spread applications: Economic models, sociology, analysis etc. Lemma 2.5. Tn (x) satisfies differential equation nk=1 xk Txk (x) = 0. Proof. It is clear that T (x) possess continuous partial derivatives. nk=1 xk Txk (x) = 0 follows immediately from Lemma 2.4. Lemma 2.6. If Txi (x) = 0 then x2i =
xi+1 +xi+2 x x xi−1 +xi−2 i−2 i−1
for i = 1, 2, . . . , n.
Proof. Proof is trivial. Lemma 2.7. T1 (x) = 2 and T2 (x) ≥ 4. Proof. T1 (x) = 2 is trivial. With x3 = x1 and x4 = x2 , T2 (x) = where x =
x1 . x2
x2 + x1 x1 + x2 1 + =2+x+ , x1 x2 x
Minimum of x +
1 x
occurs at x = 1. Thus, T2 (x) ≥ 4.
Lemma 2.8. T3 (x) ≥ 6. Proof. With x4 = x1 and x5 = x2 , T3 (x) =
1 1 x2 + x3 x3 + x4 x4 + x5 1 + + =x+ +y+ +z+ , x1 x2 x3 x y z
where x = xx12 , y = xx13 and z = (1, 1, 1). Thus, T3 (x) ≥ 6.
x2 . x3
Minimum of T3 (x) occurs at the point
Lemma 2.9. If n ≥ 4 then Tn (x) ≥ 2n. Proof. Proof is the same as Lemma 2.8. So we omit it. Lemma 2.10. Assume that xn+1 = x1 and xn+2 = x2 . If (xk + xk+1 ) ≥ 2 min{xk , xk+1 }, then T3 (x) ≥ 2n where k = 1, 2, . . . , n. Proof. If min{xk , xk+1 } = xk or xk+1 then nk=1 (xk +xk+1 ) ≥ 2n x1 x2 x3 . . . xn−1 xn . From (5) we have the result as announced. Lemma 2.11. Assumethat xn+1 = x1 and xn+2 = x2 . If (xk + xk+1 ) ≤ 2 max{xk , xk+1 }, then nk=1 F (k) ≤ 2n where k = 1, 2, . . . , n. Proof. Proof is the same as Lemma 2.10. So we omit it.
1874
3
T. Tanriverdi
Examples
We want to look at the following interesting identity [for proof, see [13, p.25]]. n−1
sin(
k=1
kπ n ) = n−1 n 2
(n = 2, 3, . . . ).
))−1 where k = 1, 2, . . . , n − 1. By using Example 3.1. Set F (k) = (sin( kπ n the identity together with (5) one gets n−1
F (k) ≥ (n − 1)(
k=1
2n−1 1 ) n−1 ≥ 2(n − 1), n
for large enough n. Since 12 ≤ 1/n1/(n−1) ≤ 1 as n varies from 2 to ∞. ))−1 where k = 1, 2, . . . , n − 1 and choose Example 3.2. Set F (k) = (sin( kπ n F (n) = 2n. By using the identity together with (5) one gets n k=1
F (k) ≥ n(
1 2n−1 2n) n ≥ 2n, n
as n varies from 2 to ∞.
References [1] H. S. Shapiro, Advanced problem 4603, Amer. Math. Monthly 61 (1954) 571. [2] P. H. Diananda, Extensions of an inequality of H. S. Shapiro, Amer. Math. Monthly 66 (1959) 489-491. [3] L. J. Mordell, On the inequality nr=1 xr /(xr+1 + xr+2 ) ≥ n2 and some others, Abh. Math. Sem. Univ. Hamburgh 22 (1958) 229-240. n n [4] L. J. Mordell, Note on the inequality r=1 xr /(xr+1 + xr+2 ) ≥ 2 , J. London Math. Soc. 37 (1962) 176-178. [5] D. Z. Djokovic, Sur une inegalite, Proc. Glasgow Math. Assoc. 6 (1963) 1-10. [6] P. Nowosad, Isoperimetric eigenvalue problems in algebras, Comm. Pure Appl. Math. 21 (1968) 401-465. [7] P. J. Bushell and A. H. Craven, On Shapiro’s cyclic inequality, Proc. Roy. Soc. Edinburgh Sect. A 26 (1975/76) 333-338.
Reverse Shapiro type inequality
1875
[8] E. K. Godunova and V. I. Levin, A cyclic sum with 12 terms, Math. Notes 19 (1976) 510-517. [9] P. J. Bushell and J. B. Mcleod, Shapiro’s cyclic inequality for even n, J. of Inequal. & Appl. 7 (2002) 331-348. [10] B. A. Troesch, On Shapiro’s cyclic inequality for n = 13, Math. Comp. 45 (1985) 199-207. [11] B. A. Troesch, The validity of Shapiro’s cyclic inequality, Math. Comp. 53 (1989) 657-664. [12] P. J. Bushell, Shapiro’s cyclic sum, Bull. London Math. Soc. 26 (1994) 564-574. [13] M. R. Spiegel, Schaum’s Outline of Theory and Problems of Complex Variables, McGraw-Hill, New York 1999. Received: March, 2012
