Hölder, Chebyshev and Minkowski Type Inequalities for Stolarsky Means

Transcription

1 Int. Journal of Math. Analysis, Vol. 4, 200, no. 34, Hölder, Chebyshev and Minkowski Type Inequalities for Stolarsky Means Zhen-Hang Yang System Division, Zhejiang Province Electric Power Test and Research Institute, Hangzhou, Zhejiang China, 3004 Abstract. In this paper, the Hölder, Chebyshev and Minkowski type inequalities for Stolarsky means are established. In addition, another new and concise proof of Minkowski type inequality for Stolarsky means is presented. Mathematics Subject Classification: Primary 26D07,26E60; Secondary 26A48, 26A5 Keywords: Hölder inequality, Chebyshev inequality, Minkowski inequality, Stolarsky means. Introduction For one-parameter means of two positive numbers x and y denoted by α(x α+ y α+ ), α 0,,x y, (α+)(x α y α ) x y, α 0,x y, (.) J(α; x, y) ln x ln y xy(ln x ln y), α,x y, x y y, x y, H. Alzer presented Chebyshev and Minkowski type inequalities in 988 (see, 2]. For generalized logarithmic means of two positive numbers x and y denoted by ( xα y α ) α α 0,,x y, α(x y) (.2) S(α; x, y) L(x, y) α 0,x y, I(x, y) α,x y, y x y,

2 688 Zhen-Hang Yang H.-E. Lou proved that the Hölder type inequality is valid in 996 (see 4]). Zh.- H. Yang proved that its Hölder, Chebyshev and Minkowski type inequalities are also true by using classical integral inequalities in 2005 (see 6]) The Stolarsky means (extended means) of positive numbers x and y is defined by (.3) E(r, s; x, y) ( ) s x r y r r s r x s y s r s, rs 0,x y, L r (x r,y r ) r 0,s0,x y, L s (x s,y s ) r 0,s 0,x y, I r (x r,y r ) r s 0,x y, G(x, y) r s 0,x y, y x y, where L(x, y) (x y) / ln(x/y),i(x, y) e (x x /y y ) x y,g(x, y) xy. In 998, L. Losonczi and Zs. Páles showed that Minkowski type inequality hold for Stolarsky means (extended means) holds, which is read as follows: Theorem (Minkowski Type Inequality for Stolarsky Mean (see 3, Theorem, 3]) ). For arbitrary pairs (x,y ), (x 2,y 2 ) R +, the following inequality (.4) E(r, s; x + x 2,y + y 2 ) ( )E(r, s; x,y )+E(r, s; x 2,y 2 ) holds iff r + s ( )3 and min(r, s) ( ) with (r, s) (2, ), (, 2). With equality iff x : y x 2 : y 2. Zh.-H. Yang studied the AA, GG, AG and GA convexity of homogeneous functions of two variables and presented simplified decisions. As applications, new simple proofs of Hölder, Chebyshev and Minkowski type inequalities for homogeneous means such as J(α; x, y) and S(α; x, y) were given (see 7, 8]). The main purpose of this paper is to investigate the GG-convexity of the Stolarsky means (extended means) by applying theorems or corollaries in 7, 8], and then Hölder and Chebyshev type inequalities for Stolarsky means (extended means) are given. Our main results are as follows: Theorem 2 (Hölder Type Inequality for Stolarsky Mean). For given p, q > 0 with p + q and arbitrary pairs (x,y ), (x 2,y 2 ) R +, inequality (.5) E(r, s; x p x q 2,yy p 2) q ( )E p (r, s; x,y )E q (r, s; x 2,y 2 ) holds iff r + s>(<)0.with equality iff x : y x 2 : y 2. Theorem 3 (Chebyshev Type Inequality for Stolarsky Mean). If (x,y ) and (x 2,y 2 ) are oppositely (similarly) ordered, then (.6) E(r, s; x x 2,y y 2 ) ( )E(r, s; x,y )E(r, s; x 2,y 2 ) if r + s>(<)0.

3 H-type, Ch-type and M-type inequalities for Stolarsky means 689 In addition, as a result of AA-convexity of the Stolarsky means (extended means) proved by L. Losonczi and Zs. Páles, another new proof of Minkowski type inequality is presented. To prove the above theorems, we need corollaries in 7] as preparations. For the sake of convenience, we read them as follows: Lemma (7, Corollary 3.2.]). Let f : D( R + R + ) R + is an n-order homogeneous and two-time differentiable function. Then for given p, q > 0 with p + q and arbitrary pairs (x,y ), (x 2,y 2 ), (x p xq 2,yp yq 2 ) D, inequality (.7) f(x p x q 2,y p y q 2) f p (x,y )f q (x 2,y 2 ) holds iff I (ln f) xy < 0. With equality iff x : y x 2 : y 2. Lemma 2 (7, Corollay 3.2.2]). Let f : D( R + R + ) R + is an n-order homogeneous and two-time differentiable function. If I (ln f(x, y)) xy < (> )0, then for arbitrary pairs (x,y ), (x 2,y 2 ) and (, ) D we have (.8) f(, )f(x x 2,y y 2 ) < (>)f(x,y )f(x 2,y 2 ) if (x,y ) and (x 2,y 2 ) are oppositely ordered. (x 2,y 2 ) are similarly ordered. It is reversed if (x,y ) and Lemma 3 (7, Corollary 3..]). Let f : D R is a one-order homogeneous function and is two-time differentiable. Then for given p, q > 0 with p + q and arbitrary pairs (x,y ), (x 2,y 2 ), (px + qx 2,py + qy 2 ) D, the following inequality (.9) f(px + qx 2,py + qy 2 ) pf(x,y )+qf(x 2,y 2 ) holds iff xyf xy < 0. With equality iff x : y x 2 : y Proofs of Theorem 2 and 3 Proof of Theorem 2. By Lemma, to prove Theorem 2, it is enough to prove that I I(r, s) 2 ln E < (>)0 for all (x, y) R + R + iff r + s>(<)0. By partial derivative calculations, for rs(r s) 0 we have ( ) s ln E r s ln x r y r (2.), r x s y s ln E ( ) E x E x rx r (2.2) x(r s) x r y sxs, r x s y s ln E ( ) E y E y ry r (2.3) y(r s) x r y + sys, r x s y ( ) s 2 ln E r 2 x r y r xy(r s) (x r y r ) s2 x s y s (2.4). 2 (x s y s ) 2

4 690 Zhen-Hang Yang ) For rs(r s) 0. Put ln x/y t, t R and use sinh x 2 (ex e x ), cosh x 2 (ex + e x ), then I can be written as ( ) r 2 I(r, s) xy(r s) 4 sinh 2 rt s 2 4 sinh 2 st (2.5) 4xyt 4xyt ( rt sinh rt + st ) rt sinh st sinh rt rt st g(rt) g(st) (g(rt)+g(st)), rt st st sinh st where g(u) u if u 0,g(0). sinh u It is easy to verify that g(u) > 0 and is even on (, ) and g (u) sinh u u cosh u sinh 2 u h(u) sinh 2 u. Since h (u) u sinh u 0, then h(u) < 0 for u (0, ), it follows that g (u) < 0 for u (0, ). Hence, and then Thus sgn g(rt) g(st) rt st g(rt) g(st) rt st sgn rt + st rt + st g( rt ) g( st ), rt st rt + st g( rt ) g( st ) sgn sgn(r + s)t. rt + st rt st sgn I(r, s) sgn g(rt) g(st) sgn (g(rt)+g(st)) sgn sgn(r + s). 4xyt rt st This shows (.5) holds iff r + s>(<)0. 2) For r 0,s(r s) 0. By (2.5) we have From ) it follows that I(0,s) : lim r 0 I(r, s) 4xyt sgn I(0,s) sgn s. 3) For s 0,r(r s) 0. Similarly, we have 4) For r s 0. By (2.5) we have sgn I(r, 0) sgn r. g(st) ( + g(st)). 0 st I(s, s) : lim r s I(r, s) 4xyt (g(st)+g(st)) g (st),

5 it follows that H-type, Ch-type and M-type inequalities for Stolarsky means 69 sgn I(s, s) sgn 4xyt sgn (g(st)+g(st)) sgn g (st) sgn s. 5) For r s 0. It is clear I(0, 0) 0. Summarizing all cases above, we see that I(r, s) < (>)0 iff s + r>(<)0. It is obvious equality holds iff x : y x 2 : y 2. This completes the proof. Proof of Theorem 3. By Lemma 2 and Theorem 2, note E(r, s;, ), we immediately obtain the result required. This proof is completed. By (2.2)-(2.4) and note 2 ln E have 3. A new Proof of Theorem Since E(r, s; x, y) is a one-order homogeneous function of variables x and y, to prove Theorem, it is enough to prove that E xy 2 E(r,s;x,y) < (>)0 for all (x, y) R + R + iff r+s ( )3 and min(r, s) ( ) with ) (r, s) (2, ), (, 2)., for rs(r s) 0we 2 E E (3.) 2 ln E +( E xy(r s) + xy(r s) 2 ( E 2 E E E 2 x E x )( E E y ) ] r 2 x r y r s2 x s y s (x r y r ) 2 (x s y s ) 2 E y ] r 2 x r y r + rs(xr y s +x s y r ) + s2 x s y s (x r y r ) 2 (x r y r )(x s y s ) (x s y s ) 2 ] r2 (r s )xryr + rs(xy)s (x r s +y r s ) s2 (r s+)x s y s. xy(r s) 2 (x r y r ) 2 (x r y r )(x s y s ) (x s y s ) 2 Substituting ln(x/y) t, t R and using sinh x 2 (ex e x ), cosh x 2 (ex + e x ), then 2 E can be expressed as E 2 E E ] r s r2 (r s ) 2rs cosh + xy(r s) 2 4 sinh 2 r 2 t s2 (r s+) 2 t 4 sinh r 2 t sinh s 2 t 4 sinh 2 s 2 t sinh 2 r 2 t sinh 2 s 2 t r 2 (r s ) sinh 2 st 4xy(r s) rs cosh r st sinh r t sinh st s2 (r s + ) sinh 2 r t]. 2 Using the half-angle formula for sinh 2 and product into sum formula for hyperbolic functions yield (3.2) 2 E E c 0A cosh rt + B cosh st + C cosh(r s)t + D],

6 692 Zhen-Hang Yang where c 0 8xy (r s) 2 sinh 2 r 2 t s sinh 2 2 t, A s(s )(r s), B r(r )(r s), C rs, D (r s) 2 (r + s) (r 2 rs + s 2 ). Next let us prove stepwise. Step. Let (3.3) g(t) :A cosh rt + B cosh st + C cosh(r s)t + D. Then 2 E < (>)0 for all (x, y) R + R + iff g (t) > (<)0. Proof. By simple calculations we get g(0) g (0) 0. Using Taylor formula, for arbitrary t R we have g(t) g (ξ) t2,ξ lie in 0 and t. 2 This shows g(t) > (<)0 for all t R iff g (t) > (<)0 for all t R. From E,c 0 > 0 and 2 E c E 0g(t) it follows that 2 E < (>)0 for all (x, y) R + R + iff g (t) > (<)0. Step is proved. Step 2. If (r + s)(r s) 0, then g (t) can be written as (3.4) g (t) rs(r s)p (t)p 2 (t), where (3.5) p (t) : cosh rt cosh st, { r(s ) + (r s) cosh(r s)t cosh st, t 0; (3.6) p 2 (t) : rs. r+s Proof. It is easy to check that (3.7) Ar 2 + Bs 2 + C(r s) 2 0. A simple derivative calculations yields g (t) Ar 2 cosh rt + Bs 2 cosh st + C(r s) 2 cosh(r s)t (3.8) Ar 2 (cosh rt cosh st)+c(r s) 2 cosh(r s)t cosh st] ] (cosh rt cosh st) Ar 2 2 cosh(r s)t cosh st + C(r s) rs(r s)(cosh rt cosh st) r(s ) + (r s) rs(r s)p (t)p 2 (t) for t 0, g (0) Ar 2 + Bs 2 + C(r s) 2 0rs(r s)p (0)p 2 (0). cosh(r s)t cosh st ]

7 H-type, Ch-type and M-type inequalities for Stolarsky means 693 Step 2 is completed. Step 3. p 2 (t) ( )0 for all t R iff p 2 (0) ( )0 and p 2 ( ) ( )0. Proof. ) we first consider the monotone of p 2 (t). Derivative calculations yield ] p cosh(r s)t cosh st 2(t) (r s) (r s) sinh(r s)t s sinh st r sinh rt s sinh st (3.9). cosh(r s)t cosh st Since x sinh x x sinh x, cosh x cosh x, so p 2 (t) can be expressed as (3.0) p 2 s (t) r t cosh(r s)t cosh st (r s)t sinh (r s)t st sinh st cosh (r s)t cosh st ] rt sinh rt st sinh st. cosh rt cosh st It is easy to prove that ( ) t sgn sinh t t 2 sinh t 2 cosh t cosh t 2 t 2 sinh t 2 t 3 sinh t (3.) 3 cosh t 2 cosh t 3 sgn(cosh t cosh t 3 ), where t,t 2,t 3 > 0 and are pairwise unequal. In fact, denote by f(x) x 2 ln(x + x 2 ). Simple derivative calculations yield f x (x) + x2 ln(x + x 2 ), f (x) x x 2 ln(x + x 2 ) ( x 2 ) 3 h (x) 2 x2 > 0, h(x) ( x 2 ) 3, and then h(x) >h() 0. It follows that f (x) > 0on(, + ). The property of convex functions implies f(x ) f(x 2 ) f(x ] 2) f(x 3 ) (3.2) > 0, x x 3 x x 2 x 2 x 3 in other words, f(x ) f(x 2 ) sgn f(x ] 2) f(x 3 ) (3.3) sgn(x x 3 ), x x 2 x 2 x 3 where x,x 2 and x 3 are pairwise unequal. Put x i cosh t i with t i > 0,i, 2, 3, then x 2 i sinh t i, ln(x i + x 2 i ) t i, consequently, (3.3) implies (3.). By (3.) and (3.0) we have ( ) r s ] sgn p 2(t) sgn sgn cosh(r s)t cosh st sgncosh (r s)t cosh rt ] t ( ) r s ] sgn sgn r(r 2s) sgn s(2r s)] (r s)(r+s) t (3.4) sgn(rs) sgn(r + s) sgn(r 2s) sgn(2r s). This shows p 2 (t) is always monotone in t R for fixed r, s.

8 694 Zhen-Hang Yang 2) Next let us observe that p 2 (0) and p 2 ( ). It is easy to verify that (3.5) p 2 (0) : lim p 2 (t) rs (r + s 3), t +0 r + s r(s ), r > s > 0;, r > 0 >s,r+ s>0; (3.6) p 2 ( ) : lim p 2 (t) t, r > 0 >s,r+ s<0; s(r ), 0 >r>s. 3) It is easy to verify that p 2 (t) is an even function on R. And then because p 2 (t) is always monotone in t R we obtain immediately that p 2 (t) 0 for all t R iff p 2 (0) 0 and p 2 ( ) 0, and p 2 (t) 0 for all t R iff p 2 (0) 0 and p 2 ( ) 0. Thus we complete step 3. Step 4. For rs(r s) 0we have ) g (t) 0 for all t R iff (3.7) 2) g (t) 0 for all t R iff (3.8) r + s 3 and min(r, s). r + s 3 and min(r, s). Proof. Without loss of generality, we assume r>s. Note (3.9) sgn(p (t)) sgn(cosh rt cosh st) sgn(rt + st) sgn(rt st) sgn(r s) sgn(r + s). Next let us consider three cases: ) For r + s > 0. By our assumption and (3.9) we obtain p (t) > 0. From (3.4) and Step 3, we see that g (t) 0,t R iff rsp 2 (t) 0 for all t R iff rsp 2 (0) 0 and rsp 2 ( ) 0. It follows from (3.5) and (3.6) that g (t) 0,t R iff Likewise g (t) 0,t R iff (3.20) r + s 3 and r>s. r + s 3 and r>s>0,s orr>0 >s. 2) For r + s<0. By our assumption and (3.9) we have p (t) < 0. From (3.4) and Step 3, we see that g (t) 0,t R iff rsp 2 (t) 0 for all t R iff rsp 2 (0) 0 and rsp 2 ( ) 0. From (3.5) and (3.6) we see that is impossible. That g (t) 0,t R iff rsp 2 (t) 0 for all t R iff rsp 2 (0) 0 and rsp 2 ( ) 0. It follows from (3.5) and (3.6) that (3.2) r>0 >sor 0 >r>s.

9 H-type, Ch-type and M-type inequalities for Stolarsky means 695 3) For r + s 0. By (3.8) we have, (3.22) g (t) C(r s) 2 cosh(r s)t cosh st] s 2 (r s) 2 cosh( 2st) cosh(st)], and then sgn(g (t)) sgn(s 2 (r s) 2 ) sgn(3s) sgn(s). Combining ) and 2) with 3), we immediately get the desired results. Step 4 is completed. Step 5. For rs(r s) 0 we have g (t) 0 for all t R iff (r, s) (2, ), (, 2). Proof. We still assume r>s. It is obvious that g (t) is not always zero for all t R if r + s 0 by (3.22), and need only consider the case of r + s 0. If (r, s) (2, ), then by a direct calculation we have g (t) rs(r s)p (t)p 2 (t) 0 for all t R. If g (t) rs(r s)p (t)p 2 (t) 0 for all t R, then p 2 (t) 0 for all t R since rs(r s) 0 and p (t) cosh rt cosh st 0. From p 2 (t) p 2 (0) p 2 (+ ) 0 we have r + s 3 0 and r(s ) 0. Solving the equalities yields (r, s) (2, ). Step 5 is completed. Step 6. For rs(r s) 0, we still have 2 E(r,s;x,y) ( )0 for all (x, y) R + R + iff r + s ( )3 and min(r, s) ( ). Proof. If r 0,s(r s) 0, then 2 lime(r, s; x, y)] E xy (0,s; x, y) r 0 2 E(r, s; x, y) lim lim (E c 0 g(t)). r 0 r 0 For r close to 0 sgn(c 0 ), by Step and 4 we have 2 E(r,s;x,y) ( )0 iff g(t) ( )0 iff g (t) ( )0 iff both r + s ( )3 and min(r, s) ( ) hold. This shows that 2 E(0,s;x,y) ( )0 iff 0 + s ( )3 and min(0,s) ( ), i.e., this Lemma is valid. Likewise, this Lemma is also true in the cases of s 0,r(r s) 0 and rs 0,r s 0 and r s 0. Step 6 is completed. Proof of Theorem. For rs(r s) 0 with (r, s) (2, ), (, 2), by Step, 4 and 5, we have E xy 2 E(r,s;x,y) < (>)0 iff r + s ( )3 and min(r, s) ( ); which is also valid for rs(r s) 0 by Step 6. Applying Lemma 3, we immediately obtain the desired results. This completes the proof.

10 696 Zhen-Hang Yang References ] H. Alzer, Üer eine einparametrige familie von Mitlewerten, Bayer. Akad. Wiss. Math. Natur. Kl. Sitzungsber., 987(988), (German) 2] H. Alzer, Üer eine einparametrige familie von Mitlewerten, II, Bayer. Akad. Wiss. Math. Natur. Kl. Sitzungsber., 988(989), (German) 3] L. Losonczi and Zs. Páles, Minkowski s inequality for two variable difference means, Proc. Amer. Math. Soc., 26(998), ] H.-W. Lou, Hölder inequalities of means, J. Ningbo Univ., 9(996), no., -7. (Chinese) 5] J. Sándor, On Lehman s inequality and electrical networks, RGMIA Research Report Collection, 8(2005), no., Article, ] Zh.-H. Yang, Hölder, Minkowski, and Chebyshev type inequalities for logarithmic mean and exponential mean, J. Xuzhou Norm. Univ. (Natural Science Edition), 23(2005), no., ] Zh.-H. Yang, Simple discriminances of convexity of homogeneous functions and applications, St. College Math., 4(2004), no. 7, 4-9. (Chinese) 8] Zh.-H. Yang, Minkowski, Hölder and Chebyshev type inequality of homogeneous functions, RGMIA Res. Rep. Coll., 8(4), Art., Received: March, 200