A refinement of Jensen‘s inequality

Vol. 17, No.1, April 2009, pp 209-214 ISSN 1222-5657, ISBN 978-973-88255-5-0, www.hetfalu.ro/octogon

209

A refinement of Jensen‘s inequality

Mih´aly Bencze and Zhao Changjian¹⁸

ABSTRACT.In this paper we give a refinement and some applications of the inequality ^1+x₂ ⁿ ≥ ^1+x₂ n

which is a particular case of the Jensen‘s inequality MAIN RESULTS

Theorem 1. Ifx≥0, n, k∈N, k≤n, then 1). x^k+x^n−k ≤1 +xⁿ

2). n 1 +xⁿ⁺¹

≥2 Pⁿ

k=1

x^n+1−k 3). 2n 1 +x+x²+...+xⁿ⁺¹

≥(n+ 2) (1 +x) 1 +x+x²+...+xⁿ 4). ^1+x₂ⁿ ≥ ^1+x+x_n+1^2+...+xn ≥ ^1+x₂ n

Proof.

1). x^k+x^n−k ≤1 +xⁿ is equivalent with x^k−1

x^n−k−1

≥0,or (x−1)²

x^k−1+x^k−1+...+x+ 1 x^n−k−1+x^n−k−2+...+x+ 1

≥0 Using 1). we have

2). n 1 +xⁿ⁺¹

= Pⁿ

k=1

1 +xⁿ⁺¹

≥ Pⁿ

k=1

x^k+x^n+1−k

= 2 Pⁿ

k=1

x^n+1−k 3). The inequality

2n 1 +x+x²+...+xⁿ⁺¹

≥(n+ 1) (1 +x) 1 +x+x²+...+xⁿ is equivalent with

1 +x+x²+...+xⁿ⁺¹

n+ 2 ≥

1 +x+...+xⁿ n+ 1

1 +x 2

.

After elementary calculus we get:

n 1 +xⁿ⁺¹

≥2 Xn k=1

x^n+1−k

18Received: 16.03.2006

2000Mathematics Subject Classification. 26D15 Key words and phrases. Jensen‘s inequality.

which result from 2).

Using 1). we get

4). (n+ 1) (1 +xⁿ) = Pⁿ

k=0

(1 +xⁿ)≥ Pⁿ

k=0

x^k+x^n−k

= 2 Pⁿ

k=0

x^k or 1 +xⁿ

2 ≥ 1 +x+x²+...+xⁿ n+ 1

The inequality ^1+x+x_n+1^2+...+xn ≥ ^1+x₂ n

we prove by induction.

Forn= 1 we have equality, ifn= 2,then (x−1)² ≥0,true. We suppose true for n,and we prove for n+ 1.But from 3) we get

1 +x+x²+...+xⁿ

n+ 1 ≥

1 +x+...+xⁿ n+ 1

1 +x 2

≥

1 +x 2

n+1

,

therefore is true for all n∈N^∗. Remark 1. The inequality

1 +xⁿ

2 ≥ 1 +x+x²+...+xⁿ

n+ 1 ≥

1 +x 2

n

is a new refinement of Jensen‘s inequality, for the function f(x) =xⁿ, n∈N.

Corollary 1.1. If x≥0 the Yn k=0

x^k+x^n−k

≤(1 +xⁿ)ⁿ Proof. From Theorem 1, point 1) we get

Yn k=0

x^k+x^n−k

≤ Yn k=0

(1 +xⁿ) = (1 +xⁿ)ⁿ Ifx= ^a_b then we have the following.

Remark 2. If a, b >0 then Yn k=0

a^kb^n−k+a^n−kb^k

≤(aⁿ+bⁿ)ⁿ

Corollary 1.2. If a, b >0 then aⁿ+bⁿ

2 ≥ aⁿ+aⁿ⁻¹b+...+abⁿ⁻¹+bⁿ

n+ 1 ≥

a+b 2

n

for all n∈N.

Proof. In Theorem 1 we getx= ^a_b.

Remark 3. If n= 2,then we obtain a problem of M. Lascu.

Corollary 1.3. If f :R→R is a convex and increasing function and g:R→R is a concave and increasing function, then

1). ^f(1)+f(x)+...+f(xⁿ)

n+1 ≥f ^1+x₂ n 2). g(1)+g(x)+...+g(xⁿ)

n+1 ≤g ^1+x₂ⁿ for all x≥0 andn∈N.

Proof. From Jensen‘s inequality and from Theorem 1 we get 1). ^f(1)+f(x)+...+f(xⁿ)

n+1 ≥f

1+x+...+xⁿ n+1

≥ ^1+x₂ n 2). g(1)+g(x)+...+g(xⁿ)

n+1 ≤g

1+x+...+xⁿ n+1

≤g ^1+x₂ⁿ

Corollary 1.4. If x, y, z >0,then Xx² ≥ 2P

x²+P xy

3 ≥

Px²+P xy

2 ≥X

xy

Proof. In Corollary 1.2 we taken= 2 so we obtain x²+y²

2 ≥ x²+xy+y²

3 ≥

x+y 2

2

= x²+ 2xy+y² 4 and

Xx² =Xx²+y²

2 ≥Xx²+xy+y²

3 = 1

3 2X

x²+X xy

≥

≥Xx²+ 2xy+y²

4 =

Px²+P xy

2 ≥X

xy

Corollary 1.5. In all triangle ABC holds 1). 2 s²−r²−4Rr

≥ ¹₃ 5s²−3r²−12Rr

≥ ¹₂ 3s²−r²−4Rr

≥

≥s²+r²+ 4Rr

2). s²−2r²−8Rr≥ ¹₃ 2s²−3r²−12Rr

≥ ¹₂ s²−r²−4Rr

≥r(4R+r) 3). (4R+r)²−2s²≥ ¹₃

2 (4R+r)²−3s²

≥ ¹₂

(4R+r)²−s²

≥s²

4). 2 8R²+r²−s²

≥ ¹₃ 5r²−8Rr+ 40R²−4s²

≥

≥ ¹₂ 3r²−8Rr+ 16R²−2s²

≥s²+r²−8Rr 5). 2

(4R+r)²−s²

≥ ¹₃

5 (4R+r)²−3s²

≥ ¹₂

3 (4R+r)²−s²

≥

≥s²+ (4R+r)²

Proof. In Corollary 1.4 we take (x, y, z)∈

(a, b, c),(s−a, s−b, s−c),(r_a, r_b, r_c), sin^{2 A}₂,sin^{2 B}₂,sin^{2 C}₂ , cos^{2 A}₂,cos^{2 B}₂,cos^{2 C}₂

Corollary 1.6. If x, y, z >0 and a, b >0, then 1). P

x²≥ ^aPx2_a+b^+bPxy ≥P xy 2). P

x²≥ P x²a

(P

xy)^b_a+b¹

≥P xy

Corollary 1.7. If x≥0 andn∈N, n≥2,then 1 +x+x²+...+xⁿ⁻¹

n

_n−1ⁿ

≤ 1 +xⁿ 2

Proof. The functionf(x) =xⁿⁿ⁻¹, x≥0 is convex, therefore from Theorem 1 point 4) we get

1 +x+x²+...+xⁿ⁻¹ n

_nⁿ₋₁

=f

1 +x+x²+...+xⁿ⁻¹ n

≤

≤ f(1) +f(x) +...+f(xⁿ)

n =

1 +xⁿⁿ⁻¹ + xⁿⁿ⁻¹2

+...+

xⁿⁿ⁻¹n−1

n ≤

≤ 1 +

xⁿⁿ⁻¹n−1

2 = 1 +xⁿ 2 Corollary 1.8. If f : (0,+∞)→R wheref(x) = √ⁿ

x, n∈N^∗, n≥2, then exist c∈n√

a+ⁿ√ b 2

;^a+b₂

,0< a≤b such that f(b)−f(a) = (b−a)f^′(c) Proof. We have f(b)−f(a) = (b−a)f^′(c) or

√n

b−√ⁿ a

b−a = 1

n√ⁿ cⁿ⁻¹

therefore

c=



 b−a n

√n

b−√ⁿ a





n n−1

=

=

√n

aⁿ⁻¹−1 + √ⁿ

aⁿ⁻²b+...+ √ⁿ

abⁿ⁻²+ √ⁿ bⁿ⁻¹ n

!_nⁿ

−1

≥

√n

b+√ⁿ a 2

!n

which follows from Corollary 1.2 fora→ √ⁿ

a, b→ √ⁿ

b,n→n−1.

In same way c=

√n

aⁿ⁻¹+ √ⁿ

aⁿ⁻²b+...+ √ⁿ

abⁿ⁻²+ √ⁿ bⁿ⁻¹ n

!_nⁿ₋₁

≤ a+b 2 which follows from Corollary 1.7 forx= qⁿ

b a.

Remark 4. Because n√a+√ⁿ b 2

n

≥√

ab, therefore for all n∈N, n≥2, c∈√

ab,^a+b₂ .

Open Question 1. Iff : (0,+∞)→R is convex and x_k>0 (k= 1,2, ..., n),then

1). ^f(x1)+f(x₂ ²⁾ ≥

f(x1)+

Pn k=1

f “_x

2 x1

”^k_n!

n+1 ≥f ^x1+x₂ ² 2). _n¹ Pⁿ

k=1

f(x_k)≥ _n(n+1)¹ Pⁿ

k=1

f(x_k) + P

cyclic

Pn k=1

f

x2

x1

_n^k!

≥f

1 n

Pn k=1

x_k

Open Question 2. Iff : (0,+∞)→R is convex andn−time differentiable, then existp_k>0 (k= 1,2, ..., n+ 1) such that for allx≥0 holds

1). ^1+x₂ⁿ ≥

Pn k=0

xn−k (n−k)!

Pn k=0

1 (n−k)!

≥ ^1+x₂ n

2). ^f(1)+f(x)₂ ≥

n+1P

k=1

pkf^(k−1)(x)

n+1P

k=1

pk

≥f ^1+x₂

REFERENCES [1] Octogon Mathematical Magazine, 1999-2009.

[2] Bencze,M., Inequalities, (manuscript), 1982.

Str. H˘armanului 6, 505600 S˘acele-N´egyfalu Jud. Bra¸sov, Romania

E-mail: benczemihaly@yahoo.com

Department of Informatics and Mathematics China Jiling University,

Hangzhou 310018, China