On Inequalities for the Exponential and Logarithmic...

Malaysian Journal of Mathematical Sciences 10(1): 23–33 (2016)

MALAYSIAN JOURNAL OF MATHEMATICAL SCIENCES

Journal homepage: http://einspem.upm.edu.my/journal

On Inequalities for the Exponential andLogarithmic Functions and Means

Bai-Ni Guo∗1 and Feng Qi2

1School of Mathematics and Informatics, Henan PolytechnicUniversity, Jiaozuo City, Henan Province, 454010, China2Department of Mathematics, College of Science, Tianjin

Polytechnic University, Tianjin City, 300387, China

E-mail: [email protected]∗Corresponding author

ABSTRACT

In the paper, the authors establish a nice inequality for the logarithmicfunction, derive an inequality for the exponential function, and recovera double inequality for bounding the exponential mean in terms of thearithmetic and logarithmic means.

Keywords: inequality, logarithmic function, exponential function, log-arithmic mean, exponential mean.

B.-N. Guo & F. Qi

1. Introduction

In (Kuang, p. 352), two inequalities

e(x+y)/2 <ex − ey

x− y<ex + ey

2

andx+ y

2<

(x− 1)ex − (y − 1)ey

ex − eyfor x 6= y are listed. The first double inequality may be derived easily fromthe Hermite-Hadamard integral inequality for the convex function et on aninterval between x and y. The second one is due to Romanian mathematicianG. Toader, but we do not know its accurate origin. Comparing the aboveinequalities, one may conjecture that

lnex − ey

x− y<

(x− 1)ex − (y − 1)ey

ex − ey< ln

ex + ey

2, x 6= y. (1)

Is the double inequality (1) really valid? The aim of this paper is to answerthe question, to confirm the validity of the double inequality (1), and to find itsequivalence, a double inequality for bounding the exponential mean in termsof the arithmetic and logarithmic means.

2. Main results

We start out from a nice inequality for the logarithmic function.

Theorem 2.1. For a > 0 and a 6= 1, we have

a− 1

ln a

(1 + ln

a− 1

ln a

)< a. (2)

Proof. Let

h(t) = ln t− (t− 1)(1− ln t+ ln |1− t| − ln | ln t|), t 6= 1.

Then straightforward computation gives

h′(t) = ln t− ln |1− t|+ ln | ln t| − 1 +t− 1

t ln t,

h′′(t) =(ln t− t+ 1)(t ln t+ 1− t)

(1− t)t2 ln2 t,

24 Malaysian Journal of Mathematical Sciences

Inequalities for Exponential and Logarithmic Functions

andlimt→1

h(t) = limt→1

h′(t) = 0.

Since1− 1

t< ln t < t− 1, t 6= 1,

it follows that h′′(t) < 0 on (0, 1) and h′′(t) > 0 on (1,∞). These imply that thefirst derivative h′(t) is decreasing on (0, 1), increasing on (1,∞), and positivefor t 6= 1. Furthermore, the function h(t) is increasing for t 6= 0, negative on(0, 1), and positive on (1,∞).

We observe that h(t) ≶ 0 for t > 0 and t 6= 1 are equivalent to

ln t ≶ (t− 1)

(1− ln t+ ln

t− 1

ln t

),

1 >t− 1

ln t

(1− ln t+ ln

t− 1

ln t

),

1

t>

1− 1/t

ln t

(1 + ln

1− 1/t

ln t

),

1

t>

1− 1/t

− ln(1/t)

[1 + ln

1− 1/t

− ln(1/t)

].

Replacing 1t by a in the last inequality leads to the inequality (2). The proof

of Theorem 2.1 is complete.

Theorem 2.2. For x 6= y, we have

lnex − ey

x− y<

(x− 1)ex − (y − 1)ey

ex − ey. (3)

Malaysian Journal of Mathematical Sciences 25

B.-N. Guo & F. Qi

Proof. The inequality (3) may be rewritten as

lnex − ey

x− y<xex − yey

ex − ey− 1,

lnex+1 − ey+1

x− y<xex − yey

ex − ey,

xex − yey

x− y>ex − ey

x− ylnex+1 − ey+1

x− y,∫ y

xet(1 + t) d t

y − x>

∫ y

xet d t

y − xln

∫ y

xet+1 d t

y − x,∫ 1

0

ex+(y−x)t[1 + x+ (y − x)t] d t >∫ 1

0

ex+(y−x)t d t ln

∫ 1

0

ex+(y−x)t+1 d t,∫ 1

0

e(y−x)t[1 + x+ (y − x)t] d t >∫ 1

0

e(y−x)t d t

[x+ 1 + ln

∫ 1

0

e(y−x)t d t

],∫ 1

0

e(y−x)t(y − x)td t >∫ 1

0

e(y−x)t d t ln

∫ 1

0

e(y−x)t d t,

(ln a)

∫ 1

0

attd t >

∫ 1

0

at d t ln

∫ 1

0

at d t,

where a = ey−x. Since the inequality (3) is symmetric between x and y, withoutloss of generality, we assume y > x which means that a > 1. Then, by a directintegration, the last inequality containing the quantity a becomes

1− a+ a ln a

ln a>a− 1

ln alna− 1

ln a,

1− a+ a ln a+ (1− a) ln(a−1ln a

)ln a

> 0,

1− a+ a ln a+ (1− a) ln(a− 1

ln a

)> 0,

a ln a+ (1− a)[1 + ln

(a− 1

ln a

)]> 0,

a ln a > (a− 1)

[1 + ln

(a− 1

ln a

)],

a >a− 1

ln a

[1 + ln

(a− 1

ln a

)].

From the inequality (2) in Theorem 2.1, the inequality (3) follows immediately.The proof of Theorem 2.2 is complete.



Theorem 2.3. For s, t > 0 and s 6= t, we have

s− tln s− ln t

<1

e

(ss

tt

)1/(s−t)

. (4)

Proof. Let ex = s and ey = t. Then the inequality (3) becomes

lns− t

ln s− ln t<

(ln s− 1)s− (ln t− 1)t

s− t

which is equivalent to

s− tln s− ln t

< exp(ln s− 1)s− (ln t− 1)t

s− t=

1

e

(ss

tt

)1/(s−t)

.

The proof of Theorem 2.3 is complete.


(x− 1)ex − (y − 1)ey

ex − ey< ln

ex + ey

2. (5)

Proof. Since the inequality (5) is symmetric with respect to x and y, withoutloss of generality, we assume y > x. Then the inequality (5) can be rearranged


B.-N. Guo & F. Qi

as

xex − yey

ex − ey< ln

ex+1 + ey+1

2,∫ y

x(1 + t)et d t∫ y

xet d t

< lnex+1 + ey+1

2,∫ 1

0[1 + x+ (y − x)t]ex+(y−x)t d t∫ 1

0ex+(y−x)t d t

< lnex+1 + ey+1

2,

1 + x+

∫ 1

0(y − x)tex+(y−x)t d t∫ 1

0ex+(y−x)t d t

< lnex+1 + ey+1

2,∫ 1

0(y − x)te(y−x)t d t∫ 1

0e(y−x)t d t

< lnex+1 + ey+1

2− (1 + x),∫ 1

0(y − x)te(y−x)t d t∫ 1

0e(y−x)t d t

< lnex+1 + ey+1

2e1+x,∫ 1

0(y − x)te(y−x)t d t∫ 1

0e(y−x)t d t

< ln1 + ey−x

2,

(ln a)∫ 1

0tat d t∫ 1

0at d t

< ln1 + a

2,

(ln a)

∫ 1

0

tat d t <

∫ 1

0

at d t ln1 + a

2,

a ln a+ 1− aln a

<a− 1

ln aln

1 + a

2,

a ln a+ 1− a < (a− 1) ln1 + a

2,

where a = ey−x. Let

H(t) = t ln t+ 1− t− (t− 1) ln1 + t

2, t > 1.

Then a straightforward computation gives

H ′(t) = ln t− t− 1

t+ 1− ln

t+ 1

2and H ′′(t) =

1− tt(t+ 1)2

< 0.

This implies that H ′(t) is decreasing and negative on (1,∞). Furthermore, thefunction H(t) is decreasing and negative on (1,∞). Hence, the inequality (5)is proved. The proof of Theorem 2.4 is complete.




1

e

(ss

tt

)1/(s−t)

<s+ t

2. (6)

Proof. Taking in (5) ex = s and ey = t figures out

(ln s− 1)s− (ln t− 1)t

s− t< ln

s+ t

2

which is equivalent to

s+ t

2> exp

(ln s− 1)s− (ln t− 1)t

s− t=

1

e

(ss

tt

)1/(s−t)

.

The proof of Theorem 2.5 is complete.

3. Remarks

Remark 3.1. The functions on both sides of the inequality (4) are respectivelycalled the logarithmic and exponential means. See Guo and Qi (2015c), Qiet al. (2014b,c,d, 2015) and plenty of references therein. Therefore, we recovera double inequality for bounding the exponential mean in terms of the arithmeticand logarithmic means.

Remark 3.2. The three inequalities (2), (3), and (4) are equivalent to eachother, although they are of different forms. Similarly, the two inequalities (5)and (6) are also equivalent to each other, although they are of different forms.

Remark 3.3. Taking in (2) a = et for t ∈ R \ {0} yields

et − 1

t

(1 + ln

et − 1

t

)< et,

that is,

lnet − 1

t<

tet

et − 1− 1.

Expanding the functions et−1t and tet

et−1 into power series results in

ln

∞∑k=1

tk−1

k!= ln

∞∑k=0

tk

(k + 1)!<

tet

et − 1− 1 =

∞∑k=1

Bk(1)

k!tk =

∞∑k=1

(−1)kBk

k!tk,


B.-N. Guo & F. Qi

where the Bernoulli polynomials Bk(x) are defined by the generating functions

zexz

ez − 1=

∞∑k=0

Bk(x)

k!zk, |z| < 2π

and the Bernoulli numbers Bk = Bk(0) = (−1)kBk(1).

According to (Comtet, 1974, pp. 140–141, Theorem A), the logarithmic poly-nomials Ln defined by

ln

(1 +

∞∑n=1

gntn

n!

)=

∞∑n=1

Lntn

n!

equal

Ln = Ln(g1, g2, . . . , gn) =

n∑k=1

(−1)k−1(k − 1)!Bn,k(g1, g2, . . . , gn−k+1),

where Bn,k denotes the Bell polynomials of the second kind which are definedby

Bn,k(x1, x2, . . . , xn−k+1) =∑

1≤i≤n,ì∈{0}∪N∑ni=1 iì=n∑ni=1 ì=k

n!∏n−k+1i=1 ì!

n−k+1∏i=1

(xii!

)ì

for n ≥ k ≥ 0. In (Guo and Qi, 2015a, Theorem 1), the formula

Bn,k

(1

2,1

3, . . . ,

1

n− k + 2

)=

n!

(n+ k)!

k∑`=0

(−1)k−`(n+ k

k − `

)S(n+ `, `)

for n ≥ k ≥ 0 was proved by two approaches, where S(n, k) denotes the Stirlingnumbers of the second kind which may be generated by

(ex − 1)k

k!=

∞∑n=k

S(n, k)xn

n!, k ∈ N.



Consequently, we obtain

lnet − 1

t= ln

∞∑k=0

tk

(k + 1)!= ln

[1 +

∞∑k=1

1

k + 1

tk

k!

]

=

∞∑n=1

[n∑

k=1

(−1)k−1(k − 1)!Bn,k

(1

2,1

3, . . . ,

1

n− k + 2

)]tn

n!

=

∞∑n=1

[n∑

k=1

(−1)k−1(k − 1)!n!

(n+ k)!

k∑`=0

(−1)k−`(n+ k

k − `

)S(n+ `, `)

]tn

n!

=

∞∑n=1

[n∑

k=1

1

k(n+kk

) k∑`=0

(−1)`+1

(n+ k

k − `

)S(n+ `, `)

]tn

n!.

As a result, it follows that

∞∑n=1

[n∑

k=1

1

k(n+kk

) k∑`=0

(−1)`+1

(n+ k

k − `

)S(n+ `, `)

]tn

n!<

∞∑n=1

(−1)nBntn

n!, t 6= 0.

The function ln et−1t may be written as

lnet − 1

t= − ln

t

et − 1= − ln

[1− t

2+

∞∑k=1

B2kt2k

(2k)!

]

= −∞∑

n=1

[n∑

k=1

(−1)k−1(k − 1)!Bn,k

(−1

2, B2, 0, B4, 0, . . . , Bn−k+2

)]tn

n!.

Remark 3.4. For given numbers b > a > 0, let

ga,b(t) =

bt − at

t, t 6= 0;

ln b− ln a, t = 0.(7)

In Guo and Qi (2009b), (Qi et al., 2009, Lemma 1), and (Guo and Qi, 2011,Theorem 2.1), the following conclusions were discovered and applied: the func-tion ga,b(t) is logarithmically convex on (−∞,∞), 3-log-convex on (−∞, 0),and 3-log-concave on (0,∞). For more and detailed information, please referto the survey article Qi et al. (2014a) and closely references therein.

Remark 3.5. In (?, Theorem 5.2), it was obtained that the logarithmic poly-nomials Ln for n ∈ N may be computed by

Ln = gn −n−1∑k=1

(n− 1

k − 1

)gn−kLk (8)


B.-N. Guo & F. Qi

and

Ln = gn + (n− 1)!

n−1∑j=1

(−1)j∑

∑ji=0 mi=n

1≤mk−1≤n−j+k−1−∑k−2

i=0 mi

1≤k≤j

mj

j∏i=0

gmi

mi!. (9)

Remark 3.6. The inequality (2) may also be rewritten as follows. For x > −1and x 6= 0, we have

x

ln(1 + x)

[1 + ln

x

ln(1 + x)

]< 1 + x, (10)

that is,x

(1 + x) ln(1 + x)

[1 + ln

x

ln(1 + x)

]< 1. (11)

The functions xln(1+x) and x

(1+x) ln(1+x) are respectively generating functions ofthe Bernoulli numbers of the second kind and the Cauchy numbers of the secondkind. See Qi (2013, 2014a,b), Qi and Zhang (2015) and closely related referencetherein.

Remark 3.7. In the papers Guo et al. (2008), Guo and Qi (2009a, 2010),Liu et al. (2008), Qi (1997, 2006), Qi et al. (2014a), Zhang et al. (2009),there have been many inequalities related to the exponential function and theirapplications.

Remark 3.8. This paper is a slightly revised version of the preprint Guo andQi (2015b).

Acknowledgements

The authors thanks the mathematicians Ming Li and Zhen-Hang Yang inChina for their valuable comments on the original version of this paper.

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