A NOTE ON THE PROOFS OF GENERALIZED RADON INEQUALITY∗ YONGTAO LIa , XIAN-MING GUc a COLLEGE OF MATHEMATICS AND COMPUTER SCIENCE, HUNAN NORMAL UNIVERSITY, CHANGSHA, 410081, P R.CHINA arXiv:1504.05874v2 [math.CA] 19 May 2015 b SCHOOL OF MATHEMATICAL SCIENCES, UNIVERSITY OF ELECTRONIC SCIENCE AND TECHNOLOGY OF CHINA, CHENGDU, 611731, P R CHINA Abstract In this note, we give different proofs of generalized Radon inequality, and then present equivalence relation between the weighted power mean inequality and Radon inequality Keywords: Bergstrăom inequality, Radon inequality, Jesen inequality MSC 2010: 26D15 Introduction The well-known Bergstrăom inequality (see e.g [13] and references therein) says that if xk , yk are real numbers and yk > for all k n, then x21 x22 x2n + + ···+ y1 y2 yn (x1 + x2 + · · · + xn )2 y1 + y2 + · · · + yn (1.1) and equality holds if and only if yx11 = yx22 = · · · = yxnn Some generalizations of the inequality (1.1) can be found in [4,5] Actually, the following Radon inequality (1.2) is just a direct consequence: If b1 , b2 , , bn are positive real numbers and a1 , a2 , , an, m are nonnegative real numbers, then am+1 am+1 am+1 n + + · · · + bm1 bm2 bmn (a1 + a2 + · · · + an )m+1 (b1 + b2 + · · · + bn )m (1.2) When m = 1, (1.2) is (1.1) For more details on Radon inequality (1.2) , the readers can refer to [6–8, pp 1351] In fact, it is easy to prove that (1.1) is equivalent to the Cauchy-Buniakovski-Schwarz inequality (see [9, pp 34-35, theorem 1.6.1]) as follows: if a1 , , an , b1 , , bn are nonnegative real numbers, then 2 n n n ak bk ak bk k=1 k=1 k=1 a Corresponding author Email addresses: liyongtaosx@outlook.com (Yongtao Li); guxianming@live.cn (Xian-Ming Gu) Recently, Mortici obtained a refinement of (1.2) in [10, pp 321, theorem 2]: am+1 am+1 am+1 n + + · · · + m m b1 b2 bmn (a1 + a2 + · · · + an )m+1 (b1 + b2 + · · · + bn )m (ai + a j )m−1 (bi a j − b j )2 + m · max i< j n bi b j (bi + b j )m In [13, Theorem 1], Yang gave the generalized Radon inequality as follows: if a1 , a2, , an are nonnegative real numbers and b1 , b2 , , bn are positive real numbers, then for r 0, s and r s + 1, ar1 ar2 arn (a1 + a2 + · · · + an )r (1.3) + + ···+ s b1s b2s bn nr−s−1 (b1 + b2 + · · · + bn )s The Weighted Power Mean inequality (see [11, pp 111-112, theorem 10.5], [6, pp 1215] and [12] for details) is as follows: if x1 , x2 , , xn are nonnegative real numbers and p1 , p2 , , pn are positive real numbers, then for r s > 0, we have p1 xr1 + p2 xr2 + · · · + pn xrn p1 + p2 + · · · + pn r p1 x1s + p2 x2s + · · · + pn xns p1 + p2 + · · · + pn s (1.4) In this paper, we give three different proofs and the applications of generalized Radon inequality (1.3), and then demonstrate the equivalence relation between the weighted power mean inequality(1.4) and Radon inequality(1.2) Main results In this section, we first give three different methods of proving generalized Radon inequality (1.3), to read for convenience, we state the result by the following theorem What’s more, we will prove equivalence relation between the weighted power mean inequality(1.4) and Radon inequality(1.2) Theorem If a1 , a2 , , an are nonnegative real numbers and b1 , b2 , , bn are positive real numbers, then for r 0, s and r s + 1, ar1 ar2 arn + + · · · + b1s b2s bns (a1 + a2 + · · · + an )r nr−s−1 (b1 + b2 + · · · + bn )s (2.1) Proof The First Proof According to (1.2), we have s+1 r n ark bs k=1 k Since r s+1 = r s+1 r s+1 ak k=1 −1 r r r s+1 a1s+1 + a2s+1 + · · · + ans+1 (b1 + b2 + · · · + bn )s bks k=1 1, then n aks+1 n (2.2) Thus, we get n = r aks+1 r k=1 s+1 −1 r (a1 + a2 + · · · + an ) s+1 r (1 + + · · · + 1) s+1 −1 (2.3) By(2.2) and (2.3), we have ar1 ar2 arn + + ···+ s b1s b2s bn (a1 + a2 + · · · + an )r nr−s−1 (b1 + b2 + · · · + bn )s Thus, (2.1) is obtained The Second Proof Let the concave function f : (0, +∞) → R be lnx We observe that the Weighted Jensen inequality: for q1 , q2 , q3 ∈ [0, 1] with q1 + q2 + q3 = and positive real numbers x1 , x2 , x3 , then we have q1 f (x1) + q2 f (x2 ) + q3 f (x3 ) f (q1 x1 + q2 x2 + q3 x3 ), and the equality holds if and only if x1 = x2 = x3 We denote ar1 ar2 arn −1 Un (a) = ( s + s + · · · + s ) b1 b2 bn and Hn (b) = (b1 + b2 + · · · + bn )−1 ar Consider x1 = bks Un (a), x2 = bk Hn (b), x3 = (n)−1 and q1 = 1r , q2 = rs , q3 = k r s + So we have s r−s−1 (q3 s 0, since r−s−1 ak (Un (a)) r · (Hn (b)) r · (n−1 ) r ark r−s−1 s · s Un (a) + · bk Hn (b) + · r bk r r n Summing over k (k = 1, 2, , n), we obtain n s k=1 n k=1 ak (Un (a)) r · (Hn (b)) r · (n−1 ) r−s−1 r r − s − −1 s ark · s Un (a) + · bk Hn (b) + · n = r bk r r The inequality (2.1) follows Before showing the third proof of theorem 1, we present a Lemma as follows Lemma If a1 , a2 , , an , b1 , b2, , bn are nonnegative real numbers and λ1 , λ2 , .,λn are nonnegative real numbers such that λ1 + λ2 + · · · + λn = 1, then n n aλk k + k=1 n bλk k (ak + bk )λk k=1 k=1 Proof According to the weighted AM-GM Inequality, we have n k=1 ak ak + bk λk n λk k=1 ak , ak + bk (2.4) and n k=1 bk ak + bk n λk bk ak + bk λk k=1 Summing up these two inequalities, we have n n aλk k + λk (ak + bk ) k=1 k=1 n k=1 which leads to the desired result (2.4) bλk k n λk = 1, k=1 Remark A particular case b1 = b2 = · · · = bn = 1, λ1 = λ2 = · · · = λn = 1 + (a1 a2 · · · an ) n (1 + a1 )(1 + a2 ) · · · (1 + an ) n n in (2.4), is a famous inequality, called Chrystal inequality(see [6, pp 61]), so we can regard lemma as a generalization of Chrystal inequality The Third Proof Use induction on n When n = 1, the result is obvious Suppose the following inequality holds when n = m: ar1 ar2 arm + + · · · + b1s b2s bms (a1 + a2 + · · · + am )r mr−s−1 (b1 + b2 + · · · + bm ) s When n = m + 1, we need to prove the following inequality: m+1 k=1 ark = bks m k=1 ark arm+1 + s bks bm+1 arm+1 (a1 + a2 + · · · + am )r (by induction assumption) + s mr−s−1 (b1 + b2 + · · · + bm )s bm+1 (Rm (a) + = arm+1 )r s bm+1 s (S m (b) + bm+1 ) r (m + 1) r r−s−1 r (m + 1)r−s−1 (S m (b) + bm+1 )s s (Rm (a)) r (S m (b)) r m r−s−1 r ar s r ) r bm+1 + ( bm+1 s r−s−1 r r m+1 1)r−s−1 (b s (m + + · · · + bm + bm+1 ) (by a special case n = in (2.4)) = where Rm (a) = (a1 + · · · + am + am+1 )r , (m + 1)r−s−1 (b1 + · · · + bm + bm+1 )s (a1 +···+am )r r−s−1 m (b1 +···+bm )s , S m (b) = b1 + b2 + · · · + bm Thus, we have ar1 ar2 arn + + · · · + b1s b2s bns (a1 + a2 + · · · + an )r , nr−s−1 (b1 + b2 + · · · + bn )s which leads to (2.1) Theorem The Radon Inequality (1.2) is equivalent to the Weighted Power Mean Inequality (1.4) Proof =⇒ By Radon inequality and y1 , y2 , , yn ∈ [0, +∞), we have r r r s r s (p1 y1 ) s r s p1 y1 + p2 y2 + · · · + pn yn = r p1s −1 + (p2 y2 ) s r p2s −1 r (pn yn ) s +···+ r pns −1 r (p1 y1 + p2 y2 + · · · + pn yn ) s ≥ r (p1 + p2 + · · · + pn ) s −1 , which implies that r r r p1 y1s + p2 y2s + · · · + pn yns p1 y1 + p2 y2 + · · · + pn yn ≥ p1 + p2 + · · · + pn p1 + p2 + · · · + pn r s (2.5) Supposing yk = xks for all xk ≥ (k = 1, 2, , n) in (2.5), we can obtain p1 xr1 + p2 xr2 + · · · + pn xrn p1 + p2 + · · · + pn ⇐= Let pk = bk , xk = ak bk r p1 x1s + p2 x2s + · · · + pn xns ≥ p1 + p2 + · · · + pn s and r = m + 1(m ≥ 0), s = in (1.4) Then, we have am+1 am+1 am+1 1 n m + m +···+ b1 + b2 + · · · + bn b1 b2 bmn m+1 ≥ a1 + a2 + · · · + an b1 + b2 + · · · + bn Thus, by (1.4) the Radon inequality (1.2) is achieved Corollary If a1 , a2 , , an , b1, b2 , , bn are positive real numbers, m am+1 am+1 am+1 n + + · · · + bm1 bm2 bmn Proof Since m −1, so (−m − 1) −1, then (a1 + a2 + · · · + an )m+1 (b1 + b2 + · · · + bn )m (2.6) 0, according to inequality (1.2), we have am+1 b−m am+1 b−m am+1 b−m 1 n n = + + · · · + + + · · · + −m−1 −m−1 −m−1 bm1 bm2 bmn a a1 a2 n (b1 + b2 + · · · + bn )−m (a1 + a2 + · · · + an )−m−1 Thus, we get inequality (2.6) Corollary If a1 , a2 , , an , b1, b2 , , bn are positive real numbers, r and s are nonpositive real numbers such that r s + , then ar1 ar2 arn + + · · · + b1s b2s bns (a1 + a2 + · · · + an )r nr−s−1 (b1 + b2 + · · · + bn )s (2.7) Proof For r obtain 0, s 0, we have that −s −r + 1, −r 0, −s By inequality (2.1), we b−s b−s ar1 ar2 b−s arn = −r + −r + · · · + n−r s + s + ···+ s b1 b2 bn a1 a2 an (b1 + b2 + · · · + bn )−s n−s−(−r)−1 (a1 + a2 + · · · + an )−r (a1 + a2 + · · · + an )r = r−s−1 (b1 + b2 + · · · + bn )s n So, inequality (2.7) holds Corollary If a1 , a2, , an , c1 , c2 , , cn are positive real numbers, and m is real numbers such that m > or m −1, then an a1 a2 + + ···+ c1 c2 cn (a1 + a2 + · · · + an )m+1 1 (2.8) m a1 c1m + a2 c2m + · · · + an cnm Proof In inequality (1.2) and (2.6), we consider bk = ak ckm for all get the inequality (2.8) Corollary If a, b ∈ R, a < b, m or m functions on [a, b] for any x ∈ [a, b], then a m+1 ( f (x)) dx (g(x))m n and then we −1, f, g : [a, b] → (0, +∞) are integrable m+1 b b k f (x)dx a b a m (2.9) g(x)dx , and ξk ∈ [xk−1 , xk ], k ∈ {1, 2, , n} By inequality Proof Letting n ∈ N, x0 = a, xk = a+k b−a n (1.2) and (2.6), we get m+1 n n k=1 f (ξk ) ( f (ξk ))m+1 (g(ξk ))m k=1 m n g(ξk ) k=1 It results that f m+1 , ∆n , ξk σ gm σ f m+1 , ∆n , ξk σ (gm , ∆n , ξk ) , m+1 m+1 where σ fgm , ∆n , ξk is the corresponding Riemann sum of function fgm , of ∆n = (x0 , x1 , , xn ) division and the intermediate ξk points By passing to limit in ineuqality above, when n tends to infinity, the inequality(2.9) follows Corollary If a, b ∈ R, a < b, rs 0, r functions on [a, b] for any x ∈ [a, b], then s + 1, f, g : [a, b] → (0, +∞) are integrable b b a r ( f (x)) dx (g(x))s a r f (x)dx b (b − a)r−s−1 a (2.10) s g(x)dx Proposition 10 Show that if a, b, c are the lengths of the sides of a triangle and 2S = a+b+c, then an bn cn + + b+c c+a a+b n−2 S n−1 , n (2.11) Proof When n = 1, the result (2.11) is Nesbitt inequality(see [9, p 16, example 1.4.8] or [11, p 2, exercise 1.3]) For n 2, by (2.1), we have bn cn an + + b+c c+a a+b (a + b + c)n 3n−1−1 (b + c + c + a + a + b) = n−2 S n−1 Proposition 11 Let a1 , a2 , , an be positive real numbers such that a1 + a2 + · · · + an = s and p q + 1 Prove that n k=1 akp (s − ak )q s p−q (n − 1)q n p−q−1 Proof By applying the inequality (2.1), the inequality above is easily obtained Proposition 12 Let x, y, and z be positive real numbers such that xyz = Prove that y3 z3 x3 + + (1 + y)(1 + z) (1 + z)(1 + x) (1 + x)(1 + y) Proof According to the generalized Radon inequality (2.1), we obtain y3 z3 x3 + + (1 + y)(1 + z) (1 + z)(1 + x) (1 + x)(1 + y) (x + y + z)3 ((1 + y)(1 + z) + (1 + z)(1 + x) + (1 + x)(1 + y)) (x + y + z)3 = + 6(x + y + z) + 3(xy + yz + zx) (by a general inequality 3(xy + yz + zx) (x + y + z)2 ) ≥ (x + y + z)3 + 6(x + y + z) + (x + y + z)2 √ Since x + y + z 3 xyz = 3, so it is easy to prove can be found in [9, pp 139-140] (x+y+z)3 9+6(x+y+z)+(x+y+z)2 , another solutions References [1] K.Y Fan, Generalization of Bergsrrăom inequality, Amer Math Monthly 66 (1959), 153154 [2] H Bergstrăom, A triangle inequality for matrices, Den Elfte 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