A note on the proofs of generalized radon inequality
aa r X i v : . [ m a t h . C A ] M a y A note on the proofs of generalized Radon inequality
Yongtao Li , Xian-Ming Gu † , , Jianci Xiao School of Mathematics and Statistics, Central South University,Changsha, Hunan, 410083, P.R. ChinaSchool of Mathematical Sciences, University of Electronic Science and Technology of China, Chengdu,Sichuan, 611731, P.R. ChinaSchool of Mathematical Sciences, Zhejiang University,Hangzhou, Zhejiang, 310027, P.R. China
Abstract
In this paper, we introduce and prove the generalizations of Radon inequality. The proofsin the paper unify and are simpler than those in former work. Meanwhile, we also findmathematical equivalences among the Bernoulli inequality, the weighted AM-GM inequality,the H¨older inequality, the weighted power mean inequality and the Minkowski inequality.Finally, a series of the applications are shown in this note.
Keywords:
Bergstr¨om inequality, Radon inequality, Weighted power mean inequality,Equivalence, H¨older inequality.
1. Introduction
The well-known Bergstr¨om inequality (see e.g. [1, 2, 3]) says that if x k , y k are real numbersand y k > ≤ k ≤ n , then x y + x y + · · · + x n y n ≥ ( x + x + · · · + x n ) y + y + · · · + y n (1.1)and the equality holds if and only if x y = x y = · · · = x n y n . E-mail address: [email protected] † Corresponding author. E-mail address: [email protected] E-mail address: [email protected] submitted to arXiv May 12, 2016 ome generalizations of the inequality (1.1) can be found in [4, 5]. Actually, the followingRadon inequality (1.2) is just a direct consequence: If b , b , . . . , b n are positive real numbersand a , a , . . . , a n , m are nonnegative real numbers, then a m +11 b m + a m +12 b m + · · · + a m +1 n b mn ≥ ( a + a + · · · + a n ) m +1 ( b + b + · · · + b n ) m . (1.2)When m = 1, (1.2) reduces to (1.1). For more details about Radon inequality (1.2), thereaders can refer to [6, pp. 1351] and [7, 8, 10]. In fact, it is not hard to prove that (1.1) isequivalent to the Cauchy-Buniakovski-Schwarz inequality (see [9, pp. 34-35, Theorem 1.6.1])stated as follows: if a , . . . , a n , b , . . . , b n are nonnegative real numbers, then n X k =1 a k n X k =1 b k ≥ n X k =1 p a k b k ! . In [14, Theorem 1], Yang has given a generalization of Radon inequality as follows: if a , a , . . . , a n are nonnegative real numbers and b , b , . . . , b n are positive real numbers, thenfor r ≥ , s ≥ r ≥ s + 1, a r b s + a r b s + · · · + a rn b sn ≥ ( a + a + · · · + a n ) r n r − s − ( b + b + · · · + b n ) s . (1.3)The weighted power mean inequality (see [12, pp. 111-112, Theorem 10.5], [7, pp. 12-15]and [13] for details) is defined as follows: if x , x , . . . , x n are nonnegative real numbers and p , p , . . . , p n are positive real numbers, then for r ≥ s >
0, we have (cid:18) p x r + p x r + · · · + p n x rn p + p + · · · + p n (cid:19) r ≥ (cid:18) p x s + p x s + · · · + p n x sn p + p + · · · + p n (cid:19) s . (1.4)In this paper, we give three different cheaper proofs and some applications of generalizedRadon inequality (1.3), and then present equivalence relations between the weighted powermean inequality and Radon inequality. Furthermore, we summarize the equivalences amongthe weighted AM-GM inequality, the H¨older inequality, the weighted power mean inequalityand the Minkovski inequality.
2. Main results
In this section, we first give three different methods for proving the generalized Radoninequality (1.3). To read for convenience, the result obtained by Yang can be stated as thefollowing theorem. 2 heorem 2.1. If a , a , . . . , a n are nonnegative real numbers and b , b , . . . , b n are positivereal numbers, then for r ≥ , s ≥ and r ≥ s + 1 , a r b s + a r b s + · · · + a rn b sn ≥ ( a + a + · · · + a n ) r n r − s − ( b + b + · · · + b n ) s . (2.1) Proof 1.
By Radon inequality (1.2), we have n X k =1 a rk b sk = n X k =1 (cid:16) a rs +1 k (cid:17) s +1 b sk ≥ (cid:16) a rs +1 + a rs +1 + · · · + a rs +1 n (cid:17) s +1 ( b + b + · · · + b n ) s . (2.2)Note that r ≥ s + 1 ≥
1, then rs +1 − ≥
0. Using Radon inequality again, we get that n X k =1 a rs +1 k = n X k =1 a rs +1 k rs +1 − ≥ ( a + a + · · · + a n ) rs +1 (1 + 1 + · · · + 1) rs +1 − . (2.3)According to inequalitis (2.2) and (2.3), we clearly have a r b s + a r b s + · · · + a rn b sn ≥ ( a + a + · · · + a n ) r n r − s − ( b + b + · · · + b n ) s . Therefore, the desired result (2.1) is obtained.
Proof 2.
Let the concave function f : (0 , + ∞ ) → R be ln x . We observe that the weightedJensen inequality: for q , q , q ∈ [0 ,
1] with q + q + q = 1 and positive real numbers x , x , x , then we have q f ( x ) + q f ( x ) + q f ( x ) ≤ f ( q x + q x + q x ) , and the equality holds if and only if x = x = x . We denote U n ( a ) = (cid:18) a r b s + a r b s + · · · + a rn b sn (cid:19) − and H n ( b ) = ( b + b + · · · + b n ) − . Consider x = a rk b sk U n ( a ) , x = b k H n ( b ) , x = n and q = r , q = sr , q = r − s − r (observe that q ≥ r ≥ s + 1). So we have a k ( U n ( a )) r · ( H n ( b )) sr · (cid:18) n (cid:19) r − s − r ≤ r · a rk b sk U n ( a ) + sr · b k H n ( b ) + r − s − r · n . k ( k = 1 , , . . . , n ), we obtain n X k =1 a k ( U n ( a )) r · ( H n ( b )) sr · (cid:18) n (cid:19) r − s − r ≤ n X k =1 (cid:18) r · a rk b sk U n ( a ) + sr · b k H n ( b ) + r − s − r · n (cid:19) = 1 . The required inequality (2.1) follows.For many numerical inequalities, the induction is some times a useful method used toestablish a given statement for all natural numbers. We now give the third proof of Theorem2.1 by mathematical induction. To state this proof clearly, let us start with the followinglemma.
Lemma 2.1. If a , a , . . . , a n , b , b , . . . , b n are nonnegative real numbers and λ , λ , . . . , λ n are nonnegative real numbers such that λ + λ + · · · + λ n = 1 , then n Y k =1 a λ k k + n Y k =1 b λ k k ≤ n Y k =1 ( a k + b k ) λ k . (2.4) Proof of lemma 2.1.
According to the weighted AM-GM inequality, we have n Y k =1 (cid:18) a k a k + b k (cid:19) λ k ≤ n X k =1 λ k (cid:18) a k a k + b k (cid:19) , Similarly, we get n Y k =1 (cid:18) b k a k + b k (cid:19) λ k ≤ n X k =1 λ k (cid:18) b k a k + b k (cid:19) . Summing up these two inequalities, we have n Y k =1 a k + b k ) λ k " n Y k =1 a λ k k + n Y k =1 b λ k k ≤ n X k =1 λ k = 1 , which leads to the desired result (2.4). Remark 2.1.
A particular case b = b = · · · = b n = 1 , λ = λ = · · · = λ n = n in (2.4)yields (1 + a )(1 + a ) · · · (1 + a n ) ≥ h a a · · · a n ) n i n , which is a famous inequality, called Chrystal inequality(see[7, pp. 61]), so we can view lemma2.1 as a generalization of Chrystal inequality. roof 3. Use induction on n . When n = 1, the result is obvious. Assume that (2.1) is truefor n = m , that is a r b s + a r b s + · · · + a rm b sm ≥ ( a + a + · · · + a m ) r m r − s − ( b + b + · · · + b m ) s . When n = m + 1, we need to prove the following inequality: m +1 X k =1 a rk b sk = m X k =1 a rk b sk + a rm +1 b sm +1 ≥ ( a + a + · · · + a m ) r m r − s − ( b + b + · · · + b m ) s + a rm +1 b sm +1 (by induction assumption)= (cid:20)(cid:16) R m ( a ) + a rm +1 b sm +1 (cid:17) r (cid:0) S m ( b ) + b m +1 (cid:1) sr ( m + 1) r − s − r (cid:21) r ( m + 1) r − s − ( S m ( b ) + b m +1 ) s ≥ h(cid:0) R m ( a ) (cid:1) r (cid:0) S m ( b ) (cid:1) sr m r − s − r + (cid:0) a rm +1 b sm +1 (cid:1) r b sr m +1 r − s − r i r ( m + 1) r − s − ( b + · · · + b m + b m +1 ) s (by a special case n = 3 in (2.4))= ( a + · · · + a m + a m +1 ) r ( m + 1) r − s − ( b + · · · + b m + b m +1 ) s , where R m ( a ) = ( a + ··· + a m ) r m r − s − ( b + ··· + b m ) s , S m ( b ) = b + b + · · · + b m . Thus, inequality (2.1) holdsfor n = m + 1, so the proof of the induction step is complete.In the next theorem, we will prove equivalence relation between the weighted powermean inequality and Radon inequality, which is partly motivated by a slight observation ofinequality (2.3). Theorem 2.2.
The Radon inequality (1.2) is equivalent to the weighted power mean inequal-ity (1.4) . Proof . = ⇒ By the Radon inequality (1.2) and y , y , . . . , y n ∈ [0 , + ∞ ) , we have p y rs + p y rs + · · · + p n y rs n = ( p y ) rs p rs − + ( p y ) rs p rs − + · · · + ( p n y n ) rs p rs − n ≥ ( p y + p y + · · · + p n y n ) rs ( p + p + · · · + p n ) rs − . which means that p y rs + p y rs + · · · + p n y rs n p + p + · · · + p n ≥ (cid:18) p y + p y + · · · + p n y n p + p + · · · + p n (cid:19) rs . (2.5)5et y k = x sk for all x k ≥ k = 1 , , . . . , n ) in (2.5). Thus, we can obtain the followingweighted power mean inequality (1.4) (cid:18) p x r + p x r + · · · + p n x rn p + p + · · · + p n (cid:19) r ≥ (cid:18) p x s + p x s + · · · + p n x sn p + p + · · · + p n (cid:19) s . ⇐ = Let p k = b k , x k = a k b k and r = m + 1( m ≥ , s = 1 in (1.4). Then, we have (cid:20) b + b + · · · + b n (cid:18) a m +11 b m + a m +12 b m + · · · + a m +1 n b mn (cid:19)(cid:21) m +1 ≥ a + a + · · · + a n b + b + · · · + b n , which implies that the Radon inequality (1.2) is achieved. Theorem 2.3.
The following inequalities are equivalent:(i) Bernoulli inequality,(ii) the weighted AM-GM inequality,(iii) H¨older inequality,(iv) the weighted power mean inequality,(v) Minkovski inequality,(vi) Radon inequality.
Proof .
The equivalence between (iv) and (vi) is given in Theorem 2.2, the equivalence among(i), (iii) and (vi), one can find in [11] as well as (ii), (iii) and (iv) in [15], the equivalencebetween (iii) and (v) is shown in [16].
Corollary 2.1. If a , a , . . . , a n , b , b , . . . , b n are positive real numbers, then for m ≤ − ,the following inequality holds a m +11 b m + a m +12 b m + · · · + a m +1 n b mn ≥ ( a + a + · · · + a n ) m +1 ( b + b + · · · + b n ) m . (2.6) Proof .
Since m ≤ −
1, thus by the inequality (1.2), we have a m +11 b m + a m +12 b m + · · · + a m +1 n b mn = b − m a − m − + b − m a − m − + · · · + b − mn a − m − n ≥ ( b + b + · · · + b n ) − m ( a + a + · · · + a n ) − m − . The inequality (2.6) holds. 6 orollary 2.2. If a , a , . . . , a n , b , b , . . . , b n are positive real numbers, then for nonpositivereal numbers r, s such that r ≥ s + 1 , we have a r b s + a r b s + · · · + a rn b sn ≥ ( a + a + · · · + a n ) r n r − s − ( b + b + · · · + b n ) s . (2.7) Proof .
For r ≤ , s ≤
0, the inequalities − s ≥ − r + 1 , − r ≥ , − s ≥ a r b s + a r b s + · · · + a rn b sn = b − s a − r + b − s a − r + · · · + b − sn a − rn ≥ ( b + b + · · · + b n ) − s n − s − ( − r ) − ( a + a + · · · + a n ) − r = ( a + a + · · · + a n ) r n r − s − ( b + b + · · · + b n ) s . So, the inequality (2.7) holds.
Corollary 2.3. If a , a , . . . , a n , c , c , . . . , c n are positive real numbers, and m is real num-bers such that m > or m ≤ − , then a c + a c + · · · + a n c n ≥ ( a + a + · · · + a n ) m +1 (cid:18) a c m + a c m + · · · + a n c m n (cid:19) m . (2.8) Proof .
Consider b k = a k c m k for all 1 ≤ k ≤ n in the inequality (1.2) and (2.6). Thus, weobtain the inequality (2.8). Corollary 2.4. If a, b ∈ R , a < b, m ≥ or m ≤ − , f, g : [ a, b ] → (0 , + ∞ ) are integrablefunctions on [ a, b ] for any x ∈ [ a, b ] , then Z ba ( f ( x )) m +1 ( g ( x )) m dx ≥ (cid:16)R ba f ( x ) dx (cid:17) m +1 (cid:16)R ba g ( x ) dx (cid:17) m . (2.9) Proof .
Letting n ∈ N + , x k = a + k b − an , k ∈ { , , . . . , n } and ξ k ∈ [ x k − , x k ]. By inequality(1.2) and (2.6), we get n X k =1 ( f ( ξ k )) m +1 ( g ( ξ k )) m ≥ (cid:18) n P k =1 f ( ξ k ) (cid:19) m +1 (cid:18) n P k =1 g ( ξ k ) (cid:19) m .
7t results that σ (cid:18) ( f ( x )) m +1 ( g ( x )) m , ∆ n , ξ k (cid:19) ≥ (cid:2) σ ( f ( x ) , ∆ n , ξ k ) (cid:3) m +1 (cid:2) σ ( g ( x ) , ∆ n , ξ k ) (cid:3) m , where σ ( f ( x ) , ∆ n , ξ k ) is the corresponding Riemann sum of function f ( x ), of ∆ n = ( x , x , . . . , x n )division and the intermediate ξ k points. By passing to limit in ineuqality above, when n tendsto infinity, the inequality(2.9) follows. Corollary 2.5. If a, b ∈ R , a < b, rs ≥ , r ≥ s + 1 , f, g : [ a, b ] → (0 , + ∞ ) are integrablefunctions on [ a, b ] for any x ∈ [ a, b ] , then Z ba ( f ( x )) r ( g ( x )) s dx ≥ (cid:16)R ba f ( x ) dx (cid:17) r ( b − a ) r − s − (cid:16)R ba g ( x ) dx (cid:17) s . (2.10) Proposition 2.1.
Show that if a, b, c are the lengths of the sides of a triangle and S = a + b + c , then a n b + c + b n c + a + c n a + b ≥ (cid:18) (cid:19) n − S n − , n ≥ . (2.11) Proof .
When n = 1, the result (2.11) is Nesbitt inequality(see [9, p. 16, example 1.4.8] or[12, p. 2, exercise 1.3]). For n ≥
2, by (2.1), we have a n b + c + b n c + a + c n a + b ≥ ( a + b + c ) n n − − ( b + c + c + a + a + b )= (cid:18) (cid:19) n − S n − . Proposition 2.2.
Let a , a , . . . , a n be positive real numbers such that a + a + · · · + a n = s and p > q + 1 > . Prove that n X k =1 a pk ( s − a k ) q ≥ s p − q ( n − q n p − q − . Proof .
By applying the inequality (2.1), the inequality above is easily obtained.
Proposition 2.3.
Let x, y , and z be positive real numbers such that xyz = 1 . Then x (1 + y )(1 + z ) + y (1 + z )(1 + x ) + z (1 + x )(1 + y ) ≥ . roof . By the generalized Radon inequality (2.1), we obtain x (1 + y )(1 + z ) + y (1 + z )(1 + x ) + z (1 + x )(1 + y ) ≥ ( x + y + z ) y )(1 + z ) + (1 + z )(1 + x ) + (1 + x )(1 + y ))= ( x + y + z ) x + y + z ) + 3( xy + yz + zx )(by a general inequality 3( xy + yz + zx ) ≤ ( x + y + z ) ) ≥ ( x + y + z ) x + y + z ) + ( x + y + z ) . Since x + y + z ≥ √ xyz = 3, it is easy to prove that ( x + y + z ) x + y + z )+( x + y + z ) ≥ . Anotherproof can be found in [9, pp. 139-140]. ReferencesReferences [1] K.Y. Fan,
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