A NOTE ON THE PROOFS OF GENERALIZED RADON INEQUALITY

arXiv:1504.05874v2 [math.CA] 19 May 2015A NOTE ON THE PROOFS OF GENERALIZED RADON INEQUALITY∗ YONGTAO LIa, XIAN-MING GUc a COLLEGE OF MATHEMATICS AND COMPUTER SCIENCE, HUNAN NORMAL UNIVE

arXiv:1504.05874v2 [math.CA] 19 May 2015

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.

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 in-equality.

Keywords: Bergstr¨om inequality, Radon inequality, Jesen inequality.

MSC 2010: 26D15

1 Introduction The well-known Bergstr¨om inequality (see e.g [1–3] and references therein) says that if

x21

y1 + x

2 2

y2 + · · · + x

2

n

y n

> (x1+ x2 + · · · + x n)2

y1+ y2+ · · · + y n

(1.1)

y1 = x2

y2 = · · · = x n

y n

Some generalizations of the inequality (1.1) can be found in [4, 5] Actually, the following

a m+1

1

b m

1

m+1

2

b m

2 + · · · + a

m+1 n

b m n

> (a1+ a2 + · · · + a n)m+1

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, ,a n,b1, ,b nare 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







2

aCorresponding 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]:

a m+1

1

b m

1

m+1

2

b m

2 + · · · + a

m+1 n

b m n

> (a1+ a2 + · · · + a n)m+1

(b1+ b2 + · · · + b n)m

+ m · max 16i< j6n

b i b j (b i + b j)m

0 and r > s + 1,

a r1

b1s + a

r

2

b s2 + · · · + a

r n

b s n

> (a1+ a2 + · · · + a n)r

n r−s−1 (b1+ b2+ · · · + b n)s. (1.3) The Weighted Power Mean inequality (see [11, pp 111-112, theorem 10.5], [6, pp

p1,p2, ,p n are positive real numbers, then for r > s > 0, we have

p1 x r

1+ p2 x r

2+ · · · + p n x r

n

p1+ p2 + · · · + p n

!1r

> p1x s

1+ p2 x s

2+ · · · + p n x s

n

p1+ p2 + · · · + p n

!1s

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)

2 Main results

In this section, we first give three different methods of proving generalized Radon inequal-ity (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 1 If a1,a2, ,a n are nonnegative real numbers and b1,b2, ,b n are positive real numbers, then for r > 0, s > 0 and r > s + 1,

a r1

b1s + a

r

2

b s2 + · · · + a

r n

b s n

> (a1+ a2 + · · · + a n)r

n r−s−1 (b1+ b2 + · · · + b n)s. (2.1)

Proof The First Proof According to (1.2), we have

n

X

k=1

a r k

b s k

=

n

X

k=1



a

r s+1

k

s+1

b s k

>



a

r s+1

r s+1

r s+1

n

s+1

n

X

k=1 a

r s+1

n

X

k=1

a

r s+1

k

(a1+ a2 + · · · + a n)s+1 r

By(2.2) and (2.3), we have

a r

1

b s

1

r

2

b s

2 + · · · + a

r n

b s n

> (a1+ a2+ · · · + a n)r

n r−s−1 (b1+ b2 + · · · + b n)s. Thus, (2.1) is obtained

The Second Proof Let the concave function f : (0, +∞) → R be lnx We observe that the

r

1

b1s + a

r

2

b2s + · · · + a

r n

b s n

and

r k

b s

r > s + 1 So we have

a k (U n (a))1r · (H n (b)) s · (n−1)r−s−1 r

r · a

r k

n.

Summing over k (k = 1, 2, , n), we obtain

n

X

k=1

a k (U n (a))1r · (H n (b)) s · (n−1)r−s−1 r

6

n

X

k=1

1

r · a

r k

r · n−1

!

= 1

The inequality (2.1) follows

Before showing the third proof of theorem 1, we present a Lemma as follows

Lemma 2 If a1,a2, ,a n,b1,b2, ,b n are nonnegative real numbers and λ1, λ2, .,λ n are nonnegative real numbers such that λ1+ λ2+ · · · + λn = 1, then

n

Y

k=1

aλk

k +

n

Y

k=1

bλk

k 6

n

Y

k=1

Proof According to the weighted AM-GM Inequality, we have

n

Y

k=1

a k

a k + b k

!λk

6

n

X

k=1

a k + b k

! ,

n

Y

k=1

b k

a k + b k

!λk

6

n

X

k=1

a k + b k

!

Summing up these two inequalities, we have

n

Y

k=1

1







n

Y

k=1

aλk

k +

n

Y

k=1

bλk

k





n

X

k=1

λk = 1,

Remark 3 A particular case b1= b2 = · · · = b n= 1, λ1= λ2 = · · · = λn= 1n in (2.4),

is a famous inequality, called Chrystal inequality(see [6, pp 61]), so we can regard lemma

1 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:

a r

1

b s

1

r

2

b s

2 + · · · + a

r m

b s m

> (a1+ a2+ · · · + a m)r

m r−s−1 (b1+ b2 + · · · + b m)s.

When n = m + 1, we need to prove the following inequality:

m+1

X

k=1

a r k

b s k =

m

X

k=1

a r k

b k s + a

r m+1

b s m+1

> (a1+ a2+ · · · + a m)r

m r−s−1 (b1+ b2 + · · · + b m)s + a

r m+1

b s m+1

(by induction assumption)

=



r m+1

b s m+1

r

>



(R m (a))1r (S m (b)) s m r−s−1 r + (a

r m+1

b s m+1

s

m+11r−s−1 r

r

(by a special case n = 3 in (2.4))

r

m r−s−1 (b1+···+b m)s , S m (b) = b1+ b2+ · · · + b m Thus, we have

a r1

b1s + a

r

2

b s2 + · · · + a

r n

b s n

> (a1+ a2 + · · · + a n)r

n r−s−1 (b1+ b2 + · · · + b n)s,

Theorem 4 The Radon Inequality (1.2) is equivalent to the Weighted Power Mean Inequality

(1.4).

Proof =⇒ By Radon inequality and y1,y2, ,y n∈ [0, +∞), we have

p1y

r

1 + p2 y

r

2 + · · · + p n y

r

n = (p1y1)

r

p r

−1

1 + (p2y2)

r

p r

−1

2 + · · · + (p n y n)

r

p r

−1

n

≥ (p1y1+ p2 y2+ · · · + p n y n)

r

(p1+ p2 + · · · + p n)r−1 , which implies that

p1y r

1 + p2 y

r

2 + · · · + p n y

r

n

p1+ p2 + · · · + p n ≥ p1y1+ p2 y2+ · · · + p n y n

p1+ p2 + · · · + p n

!r

p1 x r

1+ p2 x r

2+ · · · + p n x r

n

p1+ p2 + · · · + p n

!1r

≥ p1x

s

1+ p2 x s

2+ · · · + p n x s

n

p1+ p2 + · · · + p n

!1s

⇐= Let p k = b k,x k = a k

"

1

b1+ b2 + · · · + b n

a m+11

b m1 + a

m+1

2

b m2 + · · · + a

m+1 n

b m n

!#

1

m+1

≥ a1+ a2 + · · · + a n

b1+ b2 + · · · + b n

Corollary 5 If a1,a2, ,a n,b1,b2, ,b n are positive real numbers, m 6 −1, then

a m+1

1

b m

1

m+1

2

b m

2 + · · · + a

m+1 n

b m n

> (a1+ a2 + · · · + a n)m+1

Proof Since m 6 −1, so (−m − 1) > 0, according to inequality (1.2), we have

a m+11

b m1 + a

m+1

2

b m2 + · · · + a

m+1 n

b m n

−m

1

a −m−11 + b

−m

2

a −m−12 + · · · + b

−m

n

a −m−1

n

> (b1+ b2 + · · · + b n)−m

(a1+ a2+ · · · + a n)−m−1.

Corollary 6 If a1,a2, ,a n,b1,b2, ,b n are positive real numbers, r and s are nonpositive real numbers such that r > s + 1 , then

a r

1

b s

1

r

2

b s

2 + · · · + a

r n

b s n

> (a1+ a2+ · · · + a n)r

n r−s−1 (b1+ b2 + · · · + b n)s. (2.7)

Proof For r 6 0, s 6 0, we have that −s > −r + 1, −r > 0, −s > 0 By inequality (2.1), we

obtain

a r1

b1s + a

r

2

b2s + · · · + a

r n

b s n

−s

1

a −r1 + b

−s

2

a −r2 + · · · + b

−s

n

a −r

n

n −s−(−r)−1 (a1+ a2+ · · · + a n)−r

r

n r−s−1 (b1+ b2+ · · · + b n)s.

Corollary 7 If a1,a2, ,a n,c1,c2, ,c n are positive real numbers, and m is real numbers such that m > 0 or m 6 −1, then

a1

c1 + a2

c2 + · · · + a n

c n >

(a1+ a2 + · · · + a n)m+1



a1c

1

m

1 + a2 c

1

m

2 + · · · + a n c

1

m

n

Proof In inequality (1.2) and (2.6), we consider b k = a k c

1

m

Corollary 8 If a, b ∈ R, a < b, m > 0 or m 6 −1, f, g : [a, b] → (0, +∞) are integrable

functions on [a, b] for any x ∈ [a, b], then

a

a f (x)dx

m+1

a g(x)dx

Proof Letting n ∈ N, x0= a, x k = a+k b−a n , and ξk ∈ [x k−1,x k ], k ∈ {1, 2, , n} By inequality

(1.2) and (2.6), we get

n

X

k=1

n

P

k=1

f (ξ k)

!m+1

n

P

k=1 g(ξ k)

!m

It results that

g m ,∆n, ξk

!

>

σf m+1,∆n, ξk

σ (g m,∆n, ξk) ,

Corollary 9 If a, b ∈ R, a < b, rs > 0, r > s + 1, f, g : [a, b] → (0, +∞) are integrable

functions on [a, b] for any x ∈ [a, b], then

a

a f (x)dx

r

a 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

a n

b + c +

b n

c + a +

c n

a + b >

2

3

!n−2

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

a n

b + c +

b n

c + a +

c n

a + b >

3n−1−1 (b + c + c + a + a + b)

3

!n−2

S n−1



Proposition 11 Let a1,a2, ,a n be positive real numbers such that a1+ a2 + · · · + a n = s

and p > q + 1 > 1 Prove that

n

X

k=1

a k p

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 = 1 Prove that

x3

y3

z3

3

Proof According to the generalized Radon inequality (2.1), we obtain

x3

y3

z3

(1 + x)(1 + y)

3 ((1 + y)(1 + z) + (1 + z)(1 + x) + (1 + x)(1 + y))

3

9 + 6(x + y + z) + 3(xy + yz + zx)

3

Since x + y + z > 3√3

xyz = 3, so it is easy to prove 9+6(x+y+z)+(x+y+z) (x+y+z)3 2 > 3

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