EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 13, No. 5, 2020, 1199-1211 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Special Issue Dedicated to Professor Hari M. Srivastava On the Occasion of his 80th Birthday Schur Geometric Convexity of Related Function for Holders Inequality with application Sreenivasa Reddy Perla1,∗, S Padmanabhan2, V Lokesha3 1 Department of Mathematics, The Oxford College of Engineering, Bangalore, Karnataka, India 2 Department of Mathematics, RNS Institute of Technology, Bangalore, Karnataka, India 3 Department of Mathematics, VSK University, Bellary, Karnataka, India Abstract. In this paper, we investigated the Schur geometric convexity of related function for Holders Inequality by using majorization inequality theory, giving a complete critical condition of Schur Geometrically convex function for Holders Inequality related function and some applications are established. 2020 Mathematics Subject Classifications: 26E60, 26D15, 26A51, 34K38 Key Words and Phrases: Holders Inequality, majorization inequality, schur geometric convex, schur geometric concave 1. Introduction Throughout this paper, we assume that the set of n-dimensional row vector on the real number field by Rn. Let Rn+ = {x = (x1, x2, ...xn) : xi ≥ 0, i = 1, 2...n}, By Holders inequality [2], we have n∑ l=1 rlsl ≤ ( n∑ l=1 rul ) 1 u ( n∑ l=1 svl ) 1 v (1) ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v13i5.3741 Email addresses: srireddy sri@yahoo.co.in (S. R. Perla), padmanabhanrnsit@gmail.com (S. Padmanabhan), v.lokesha@gmail.com (V. Lokesha) https://www.ejpam.com 1199 c© 2020 EJPAM All rights reserved. S. R. Perla, S. Padmanabhan, V. Lokesha / Eur. J. Pure Appl. Math, 13 (5) (2020), 1199-1211 1200 ∫ s r φ(x)ψ(x)dx ≤ (∫ r s( φ(x) )u dx ) 1 u (∫ r s( ψ(x) )v dx ) 1 v (2) Here rl ≥ 0, sl ≥ 0, u > 1, 1u + 1 v = 1. The Schur convexity of functions relating to special means is a very significant research subject and has attracted the interest of many mathematicians.There are numerous ar- ticles written on this topic in recent years; (see [3] , [6]) and the references therein. As supplements to the Schur convexity of functions, the Schur geometrically convex functions and Schur harmonically convex functions were investigated by Zhang and Yang ([15], [13]), Chu, Zhang and Wang [14], Shi and Zhang ([8], [7]) , Meng, Chu and Tang [4] , Zheng, Zhang and Zhang [17]. These properties of functions have been found to be useful in discovering and proving the inequalities for special means (see [1] - [2], [11],[12]). Dong-Sheng Wang, Chun - Ru Fu and Huan-Nan Sh [10] investigated the Schur convex- ity about related function of Holders inequality by using majorization inequality theory . This result gives a full essential condition of Schur convexity for Holders inequality related function, reached sharpen type of Holders inequality Under certain conditions and new inequalities for Stolarsky mean estabilsihed. This paper motivates us to investigate Schur geometric convexity about related function of Holders inequality by using majorization inequality theory. 2. Preliminaries To estabilish our main results, we need the following definitions and lemmas. Definition 1. [[3], [9]] .Consider two arbitrary n-tuple elements λ, µ ∈ Rn λ = (λ1, λ2, ..., λn) and µ = (µ1, µ2, ..., µn) ∈ Rn . (i) For the arrangements of λ and µ in descending order of the form if t∑ p=1 λ[p] ≤ t∑ p=1 µ[p] for 1 ≤ t ≤ n− 1, λ is said to by majorized by µ, (in icon λ ≺ µ) and n∑ p=1 λ[p] = n∑ p=1 µ[p], where λ[1] ≥ · · · ≥ λ[n] and µ[1] ≥ · · · ≥ µ[n] (ii) Let Ψ ⊆ Rn (n ≥ 2) p = 1, 2, · · · , n λ ≥ µ means λp ≥ µp . The function ω : Ψ→R is declining if and just if −ω is escalating. S. R. Perla, S. Padmanabhan, V. Lokesha / Eur. J. Pure Appl. Math, 13 (5) (2020), 1199-1211 1201 (iii) For ζ, η ∈ [0, 1] with ζ+η = 1, Ψ ⊆ Rn is a convex set, if (ζλ1 + ηµ1, · · · , ζλn + ηµn) ∈ Ψ for all λ and µ. (iv) the function ω : Ψ → R is considered to be Schur-convex whenever λ ≺ µ on Ψ implies ω(λ) ≤ ω(µ). ω is Schur concave on Ψ if −ω is Schur convex. Lemma 1. [5]. Let ω : Ψ→ R be differentiable in Ψ0 and continuous on Ψ and Ψ ⊆ Rn be symmetric with non-empty interior Ψ0, then ω is Schur convex on Ψ if and only if ω is symmetric on Ψ and (p− q) ( ∂ω ∂p − ∂ω ∂q ) ≥ 0(≤ 0) (3) Definition 2. [8]. If (λζ1µ η 1, ...., λ ζ nµ η n)∈ Ψ for all λ and µ ∈ Ψ and λ = (λ1, λ2, ..., λn) and µ = (µ1, µ2, ..., µn) ∈ Rn+, then Ψ ⊆ Rn is identified as geometrically convex set, where ζ, η ∈ [0, 1] with ζ + η = 1. If (lnλ1, ..., lnλn) ≺ (lnµ1, ..., lnµn) on Ψ implies ω(λ) ≤ ω(µ) and Ψ ⊆ Rn+, then the function ω : Ψ→ R+ is called as Schur geometrically convex function on Ψ. Lemma 2. [8]. Let ω : Ψ→ R be differentiable in Ψ0 and continuous on Ψ and Ψ ⊆ Rn be symmetric with non-empty interior Ψ0, then ω is Schur Schur-geometrically convex (Schur-geometrically concave) function. If ω is symmetric on Ψ and (ln p− ln q) ( p ∂ω ∂p − q∂ω ∂q ) ≥ 0(≤ 0). (4) Lemma 3. [16]. (Chebyshev’s inequality) If progressions rn ≥ 0, sn ≥ 0 we have (i) When rn, sn have opposite monotonicity, then n∑ l=1 r1 n∑ l=1 sl ≥ n n∑ l=1 slrl (5) (ii) When rn, sn have same monotonicity, then n∑ l=1 rl n∑ l=l s1 ≤ n n∑ l=1 slrl (6) Lemma 4. [16]. If φ(x) is the convex (concave) function on the interval then φ ( r + s 2 ) ≤ (≥) 1 s− r ∫ r s φ(x)dx ≤ (≥) ( φ(r) + φ(s) 2 ) (7) Lemma 5. [9]. Let x = (x1, x2, x3...xn) ∈ Rn and An(x) = 1 n ∑n i=1 xi. then u = (An(x), An(x), ..., An(x))︸ ︷︷ ︸ < (x1, x2, ...xn) = x n S. R. Perla, S. Padmanabhan, V. Lokesha / Eur. J. Pure Appl. Math, 13 (5) (2020), 1199-1211 1202 Lemma 6. [2]. (Young’s inequality) Suppose r, s ≥ 0, u ≥ 1, 1u + 1 v = 1 then 1 u ru + 1 v su ≥ rs (8) Lemma 7. Suppose r, s ≥ 0, u ≥ 1, 1u + 1 v = 1 then rs ≤ 1 u (ru + su) + 1 v (rv + sv)− r2 + s2 2 (9) Lemma 8. when 1 ≥ r ≥ s ≥ 0, u ≥ v ≥ 1 then 1 u ru + 1 v sv ≤ 1 u su + 1 v rv (10) 3. Main Results In this paper, by using the principle of majorization as an example, combined with majorization inequality, the Schur-geometrically convexity of Related Function for Holder’s Inequality gives sharpening inequality of the Holders under certain conditions. Our primary outcome is as follows: Theorem 1. Let rn ≥ 0 and sn ≥ 0 be any two progressions and let u and v be two non-zero arbitrary real numbers. Let H1(r) = n∑ l=1 rlsl ≤ ( n∑ l=1 rul ) 1 u ( n∑ l=1 svl ) 1 v (11) If u ≥ 1, then H1(r) is Schur-geometric convex on R+ with r1, ..., rn and if u < 1, then H1(r) is Schur-geometric concave on R+ with r1, ..., rn. Proof. : Here H1(r) is obviously symmetric with r = r1, ..., rn on R+. Let us assume r1 > r2. Now by differentiating (11) partially with respect to r1 and r2, we get ∂H1 ∂r1 = ( n∑ l=1 rul ) 1 u −1( n∑ l=1 svl ) 1 v ru−1 1 and ∂H1 ∂r2 = ( n∑ l=1 rul ) 1 u −1( n∑ l=1 svl ) 1 v ru−1 2 Consider, 41 = (ln r1 − ln r2) ( r1 ∂H1 ∂r1 − r2 ∂H1 ∂r2 ) S. R. Perla, S. Padmanabhan, V. Lokesha / Eur. J. Pure Appl. Math, 13 (5) (2020), 1199-1211 1203 ⇒ 41 = (ln r1 − ln r2) ( n∑ l=1 rul ) 1 u −1( n∑ l=1 svl ) 1 v ( ru1 − ru2 ) It is easy to see that, when u ≥ 1, then 41 ≥ 0 and when u ≤ 1, then 41 ≤ 0. Hence, by Lemma 2, if u ≥ 1, then H1(r) is Schur-geometric convex on R+ with r1, ..., rn and if u ≤ 1, then H1(r) is Schur-geometric concave on R+ with r1, ..., rn. This completes proof of Theorem 1. Theorem 2. Let rn ≥ 0 and sn ≥ 0 be any two progressions and let u and v be two non-zero arbitrary real numbers. Let H2(s) = n 1 uAn,r ( n∑ l=1 svl ) 1 v (12) If v ≥ 1, then H2(s) is Schur-geometric convex on R+ with s1, ..., sn and if v ≤ 1, H2(s) is Schur-geometric concave on R+ with s1, ..., sn. Here An,r = 1 n n∑ l=1 rl. Proof. : Here H2(r) is obviously symmetric with s = s1, ..., sn on R+. Let us assume s1 > s2. Now by differentiating (12) partially with respect to s1 and s2, we get ∂H2 ∂s1 = n 1 uAn,r ( n∑ l=1 svl ) 1 v sv−1 1 and ∂H2 ∂s2 = n 1 uAn,r ( n∑ l=1 svl ) 1 v sv−1 2 Consider, 42 = (ln s1 − ln s2) ( s1 ∂H1 ∂s1 − s2 ∂H1 ∂s2 ) ⇒ 42 = (ln s1 − ln s2)n 1 uAn,r ( n∑ l=1 svl ) 1 v (sv1 − sv2) S. R. Perla, S. Padmanabhan, V. Lokesha / Eur. J. Pure Appl. Math, 13 (5) (2020), 1199-1211 1204 It is easy to see that,when v ≥ 1, then 42 ≥ 0 and when v ≤ 1, then 42 ≤ 0. Hence, by Lemma 2, if v ≥ 1 , then H2(s) is Schur-geometric convex on R+ with s1, ..., sn and if v ≤ 1, then H2(s) is Schur-geometric concave on R+ with s1, ..., sn. This completes proof of Theorem 2. Theorem 3. Let φ(x) and ψ(x) be two continuous functions with φ(x) > 0, ψ(x) > 0 and let ∫ s r φ(x)ψ(x)dx 6= 0, ∫ r s( φ(x) )u dx 6= 0, ∫ s r ( ψ(x) )v dx 6= 0, where u and v are arbitrary real numbers. Let H3 ( r, s ) =  [ ∫ s r ( ψ(x) )v dx∫ s r φ(x)ψ(x)dx ]u [ ∫ s r ( φ(x) )u dx∫ s r φ(x)ψ(x)dx ]v , if r 6= s ( φ(x)ψ(x) )uv−u−v , if r = s (13) Then H3 ( r, s ) is Schur- geometric concave(convex) with r, s if and only if: v ( φu(s) + φu(r) )∫ s r φ u(x)dx + u ( ψv(s) + ψv(r) )∫ s r ψ v(x)dx ≤ (≥) ( φ(s)ψ(s) + φ(r)ψ(r) ) (u+ v)∫ s r φ(x)ψ(x)dx (14) Proof. : HereH3 ( r, s ) is obviously symmetric with r = r1, r2, ..., rn and s = s1, s2, ..., sn on R+. Let us assume s > r. From (13), we have H3 ( r, s ) = [ ∫ s r ( ψ(x) )v dx∫ s r φ(x)ψ(x)dx ]u [ ∫ s r ( φ(x) )u dx∫ s r φ(x)ψ(x)dx ]v ⇒ H3 ( r, s ) = (∫ s r φ u(x)dx )v(∫ s r ψ v(x)dx )u (∫ s r φ(x)ψ(x)dx )u+v Now by differentiating this partially with respect to s and r, we get ∂H3 ∂s = v (∫ s r φ u(x)dx )v−1 φu(s) ∫ s r ( ψv(x)dx )u ∫ s r ( φ(x)ψ(x)dx )u+v (∫ s r φ(x)ψ(x)dx )2(u+v) S. R. Perla, S. Padmanabhan, V. Lokesha / Eur. J. Pure Appl. Math, 13 (5) (2020), 1199-1211 1205 + u (∫ s r ψ v(x)dx )u−1 ψv(s) ∫ s r ( φv(x)dx )u ∫ s r ( φ(x)ψ(x)dx )u+v (∫ s r φ(x)ψ(x)dx )2(u+v) − (u+ v) (∫ s r φ(x)ψ(x)dx )(u+v−1) φ(s)ψ(s) (∫ s r φ u(x)dx )v(∫ s r ψ v(x)dx )u (∫ s r φ(x)ψ(x)dx )2(u+v) ∂H3 ∂r = v (∫ s r φ u(x)dx )v−1 φu(r) ∫ s r ( ψv(x)dx )u ∫ s r ( φ(x)ψ(x)dx )u+v (∫ s r φ(x)ψ(x)dx )2(u+v) − u (∫ s r ψ v(x)dx )u−1 ψv(r) ∫ s r ( φv(x)dx )v ∫ s r ( φ(x)ψ(x)dx )u+v (∫ s r φ(x)ψ(x)dx )2(u+v) + (u+ v) (∫ s r φ(x)ψ(x)dx )(u+v−1) φ(r)ψ(r) (∫ s r φ u(x)dx )v(∫ s r ψ v(x)dx )u (∫ s r φ(x)ψ(x)dx )2(u+v) Consider, 43 = (ln s− ln r) ( s ∂H3 ∂s − r∂H3 ∂r ) This implies that, 43 = (ln s− ln r)(∫ s r φ(x)ψ(x)dx )2(u+v) [ v (∫ s r φu(x)dx )v−1(∫ s r ψv(x)dx )u × (∫ s r φ(x)ψ(x)dx )u+v( sφu(s) + rφu(x) ) + u (∫ s r ψv(x)dx )u−1 × (∫ s r φu(x)dx )v ∫ s r ( φ(x)ψ(x)dx )u+v( sψv(s) + rψv(r) ) S. R. Perla, S. Padmanabhan, V. Lokesha / Eur. J. Pure Appl. Math, 13 (5) (2020), 1199-1211 1206 −(u+v) ∫ s r ( φ(x)ψ(x)dx )u+v−1(∫ s r φu(x)dx )v(∫ s r ψv(x)dx )u × ( sφ(s)ψ(s)+rφ(r)ψ(r) )] = (ln s− ln r)(∫ s r φ(x)ψ(x)dx )2(u+v) [ v ∫ s r ( φ(x)ψ(x)dx )u+v−1(∫ s r φu(x)dx )v−1 × (∫ s r ψv(x)dx )u{∫ s r φ(x)ψ(x)dx ( sφu(s)+rφu(r) ) − (∫ s r φu(x)dx ) × ( sφ(s)ψ(s)+rφ(r)ψ(r) )} +u ∫ s r ( φ(x)ψ(x)dx )u+v−1(∫ s r φu(x)dx )v × (∫ s r ψv(x)dx )u−1{∫ s r φ(x)ψ(x)dx ( sψv(s)+rψu(r) ) − (∫ s r ψu(x)dx ) × ( sφ(s)ψ(s)+rφ(r)ψ(r) )}] = (ln s− ln r)(∫ s r φ(x)ψ(x)dx )2(u+v) (∫ s r φ(x)ψ(x)dx )u+v−1(∫ s r φu(x)dx )v−1(∫ s r ψv(x)dx )u−1 { v ∫ s r ψv(x)dx [∫ s r φ(x)ψ(x)dx ( sφu(s) + rφu(r) ) − ∫ s r φu(x)dx ( sφ(s)ψ(s) ) + rφ(r)ψ(r) ] + u ∫ s r φu(x)dx [∫ s r φ(x)ψ(x)dx ( sψu(s)+rψu(r) ) − ∫ s r ψv(x)dx ( sφ(s)ψ(s) ) +rφ(r)ψ(r) ]} Since (ln s− ln r)(∫ s r φ(x)ψ(x)dx )2(u+v) (∫ s r φ(x)ψ(x)dx )u+v−1(∫ s r φu(x)dx )v−1(∫ s r ψv(x)dx )u−1 ≥ 0 So 43 and v ∫ s r ψv(x)dx [∫ s r φ(x)ψ(x)dx ( sφu(s) + rφu(r) ) − ∫ s r φu(x)dx ( sφ(s)ψ(s) ) + rφ(r)ψ(r) ] S. R. Perla, S. Padmanabhan, V. Lokesha / Eur. J. Pure Appl. Math, 13 (5) (2020), 1199-1211 1207 + u ∫ s r φu(x)dx [∫ s r φ(x)ψ(x)dx ( sψu(s)+rψu(r) ) − ∫ s r ψv(x)dx ( sφ(s)ψ(s) ) +rφ(r)ψ(r) ] = ∫ s r φ(x)ψ(x)dx [ v ∫ s r ψv(x)dx ( sφu(s) + rφu(r) ) + u ∫ s r φv(x)dx ( sφv(s) + rψv(r) )] − ∫ s r φu(x)dx ∫ s r ψv(x)dx ( sφ(s)ψ(s) + rφ(r)ψ(r) )( u+ v ) have the same symbol. Hence, we have H3 ( r, s ) is Schur- Geometric concave (convex) with r, s, if and only if:∫ s r φ(x)ψ(x)dx [ v ∫ s r ψv(x)dx ( sφu(s) + rφu(r) ) + u ∫ s r φv(x)dx ( sφv(s) + rψv(r) )] ≤ (≥) ∫ s r φu(x)dx ∫ s r ψv(x)dx ( sφ(s)ψ(s) + rφ(r)ψ(r) )( u+ v ) ⇔ v ∫ s r ψ v(x)dx ( sφu(s) + rφu(r) ) + u ∫ s r φ v(x)dx ( sφv(s) + rψv(r) )∫ s r φ u(x)dx ∫ s r ψ v(x)dx ≤ (≥) ( sφ(s)ψ(s) + rφ(r)ψ(r) )( u+ v )∫ s r φ(x)ψ(x)dx ⇔ v ( sφu(s) + rφu(r) )∫ s r φ u(x)dx + u ( sψv(s) + rψv(r) )∫ s r ψ v(x)dx ≤ (≥) ( sφ(s)ψ(s) + rφ(r)ψ(r) ) (u+ v)∫ s r φ(x)ψ(x)dx This completes proof of Theorem 3. Corollary 1. Let φ(x) and ψ(x) be two continuous functions and let their second order derivatives exists with φ(x > 0, ψ(x) > 0), ∫ s r φ(x)ψ(x)dx 6= 0, ∫ r s (φ(x))udx 6= 0, ∫ r s ψ(x)udx 6= 0. If u, v > 1 and φ(x), ψ(x) are convex functions of opposite monotonicity and φppψ + ψppφ+ 2φpψp < 0 then H3(r, s) is Schur-geometric convex with r = r1, r2, ..., rn, and s = s1, s2, ...sn on R+. S. R. Perla, S. Padmanabhan, V. Lokesha / Eur. J. Pure Appl. Math, 13 (5) (2020), 1199-1211 1208 Corollary 2. Let φ(x) and ψ(x) be two continuous functions and let their second order derivatives exists with φ(x > 0), ψ(x) > 0, ∫ s r φ(x)ψ(x)dx 6= 0, ∫ r s (φ(x))udx 6= 0, ∫ r s ψ(x)udx 6= 0. If u, v < 0 and φ(x), ψ(x) are concave functions of opposite monotonicity then H3(r, s) is Schur-geometric concave with r = r1, r2, ..., rn, and s = s1, s2, ...sn on R+. Corollary 3. Let φ(x) and ψ(x) be two continuous functions and let their second order derivatives exists with φ(x > 0), ψ(x) > 0, ∫ s r φ(x)ψ(x)dx 6= 0, ∫ r s (φ(x))udx 6= 0, ∫ r s ψ(x)udx 6= 0. If −1 < u < 0, 0 < v < 1, u+v > 0 and φ(x), ψ(x) are concave functions of opposite mono- tonicity then H3(r, s) is Schur-geometric convex with r = r1, r2, ..., rn, and s = s1, s2, ...sn on R+. 4. Application The following applications are established by using our main results. Theorem 4. Let rn ≥ 0 and sn ≥ 0 be any two progressions and let u and v be two non-zero arbitrary real numbers. Then (i) if u ≥ 1, v ≥ 1 then( n∑ i=1 rul ) 1 u ( n∑ i=1 rvl ) 1 v ≥ ( n 1 u + 1 v ) An,r An,s . (ii) if u ≤ 1, v ≤ 1 then( n∑ i=1 rul ) 1 u ( n∑ i=1 rvl ) 1 v ≤ ( n 1 u + 1 v ) An,r An,s . Here An,r = ∑n l=1(rl) n , An,s = ∑n l=1(sl) n Proof. : (i) By Lemma 7 has a majorization inequality: (r1, r2, ..., rn) � ( r1 + r2 + r3 + ...+ rn n , ..., r1 + r2 + r3 + ...+ rn n ) S. R. Perla, S. Padmanabhan, V. Lokesha / Eur. J. Pure Appl. Math, 13 (5) (2020), 1199-1211 1209 and by Theorem 1 and by definition 1, we have if u ≥ 1, then H1(r) ≥ H1(An, r), that is( n∑ l=1 rul ) 1 u ( n∑ l=1 svl ) 1 v ≥ ( n(An,r) u ) 1 u ( n∑ l=1 svl ) 1 v = n 1 uAn,r ( n(An,s )v ) 1 v By majorization inequality, we have (s1, s2, ..., sn) � ( s1 + s2 + s3 + ...+ sn n , ..., s1 + s2 + s3 + ...+ sn n ) and by Theorem 2 and Definition 1, we have if v ≥ 1, then H2(s) ≥ H2(An, s), that is n 1 uAn,r ( n∑ l=1 svj ) 1 v ≥ n 1 uAn,r ( n(An,s) v ) 1 v = n ( 1 u + 1 v ) An,r An,s From the above relations, we have ( n∑ l=1 rul ) 1 u ( n∑ l=1 svj ) 1 v ≥ n ( 1 u + 1 v ) An,r An,s exactness. By Similar method the following inequality is also established, ( n∑ l=1 rul ) 1 u ( n∑ l=1 svj ) 1 v ≤ n ( 1 u + 1 v ) An,r An,s (15) The proof of Theorem 4 is complete. Theorem 5. Let rn ≥ 0 and sn ≥ 0 be any two progressions and let u and v be two non-zero arbitrary real numbers . Then (i) When u > 1, if 1 u + 1 v = 1 and {rn}, {sn} have the opposite of monotonicity, then ( n∑ l=1 rul ) 1 u ( n∑ l=1 svj ) 1 v ≥ nAn,r An,s≥ n∑ l=1 r1s1 (ii) When 0 < u < 1, if 1 u + 1 v = 1 and {rn},{sn} have the opposite of monotonicity, then ( n∑ l=1 rul ) 1 u ( n∑ l=1 svj ) 1 v ≤ nAn,r An,s≥ n∑ l=1 r1s1 REFERENCES 1210 Proof. : (i) When u > 1, if 1 u + 1 v = 1 and by Theorem 1, we have ( n∑ l=1 rul ) 1 u ( n∑ l=1 svj ) 1 v ≥ nAn,r An,s = An,r An,s and by Lemma 5, we have nAn,r An,s = n ∑n l=1(r1) n ∑n l=1(s1) n = ∑n l=1(r1) ∑n l=1(s1) n ≥ n ∑n l=1 r1s1 n = n∑ l=1 r1s1 From the above relations, we have( n∑ l=1 rul ) 1 u ( n∑ l=1 svj ) 1 v ≥ nAn,r An,s≥ n∑ l=1 r1s1 exactness. By Similar method the following inequality is also established( n∑ l=1 rul ) 1 u ( n∑ l=1 svj ) 1 v ≤ nAn,r An,s≥ n∑ l=1 r1s1 The proof of Theorem 5 is complete. 5. 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