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New inequalities for hyperbolic functions and their applications

Abstract

In this paper, we obtain some new inequalities in the exponential form for the whole of the triples about the four functions {1,(sinht)/t,exp(tcotht1),cosht}. Then we generalize some well-known inequalities for the arithmetic, geometric, logarithmic, and identric means to obtain analogous inequalities for their p th powers, where p>0.

MSC: 26E60, 26D07.

1 Introduction

Let sinht, cosht, and cotht be the hyperbolic sine, hyperbolic cosine, and hyperbolic cotangent, respectively. It is well known that (see [16])

1< sinh t t < e t coth t 1 <cosht
(1.1)

holds for all t0.

In the recent paper [7], we have established the following Cusa-type inequalities of exponential type for the triple {1,(sinht)/t,cosht} described as follows.

Theorem 1.1 (Cusa-type inequalities [[7], Part (i) of Theorem 1.1])

Let p4/5, and t0. Then the double inequality

(1λ)+λ ( cosh t ) p < ( sinh t t ) p <(1η)+η ( cosh t ) p
(1.2)

holds if and only if η1/3 and λ0.

On the other hand, the author of this paper [8] obtains the following inequalities of exponential type for the triple {1,exp(tcotht1),cosht}.

Theorem 1.2 ([[8], Theorem 2])

Let p>0, and t0. Then

(1) if 0<p6/5, the double inequality

α ( cosh t ) p +(1α)< e p ( t coth t 1 ) <β ( cosh t ) p +(1β)
(1.3)

holds if and only if α2/3 and β ( 2 / e ) p ;

(2) if p2, the double inequality

α ( cosh t ) p +(1α)< e p ( t coth t 1 ) <β ( cosh t ) p +(1β)
(1.4)

holds if and only if α ( 2 / e ) p and β2/3.

Next, we do the work for each of the triples {(sinht)/t,exp(tcotht1),cosht} and {1,(sinht)/t,exp(tcotht1)}, and obtain the following two new results.

Theorem 1.3 Let 0<p8/5, and t0. Then

α ( cosh t ) p +(1α) ( sinh t t ) p < e p ( t coth t 1 ) <β ( cosh t ) p +(1β) ( sinh t t ) p
(1.5)

holds if and only if α1/2 and β ( 2 / e ) p .

Theorem 1.4 Let p286/693, and t0. Then

α+(1α) e p ( t coth t 1 ) < ( sinh t t ) p <β+(1β) e p ( t coth t 1 )
(1.6)

holds if and only if β1/2 and α1.

In this paper, we shall give the elementary proofs of Theorem 1.3 and Theorem 1.4. In the last section, we apply Theorems 1.1-1.4 to obtain some new results for four classical means.

2 Lemmas

Lemma 2.1 ([911])

Let f,g:[a,b]R be two continuous functions which are differentiable on (a,b). Further, let g 0 on (a,b). If f / g is increasing (or decreasing) on (a,b), then the functions (f(x)f( b ))/(g(x)g( b )) and (f(x)f( a + ))/(g(x)g( a + )) are also increasing (or decreasing) on (a,b).

Lemma 2.2 Let t(0,+). Then the inequality

D(t)t sinh 5 t+2t sinh 3 t+ t 4 cosht sinh 4 tcosht t 3 sinh 3 t2 t 3 sinht>0

holds.

Proof Using the power series expansions of the functions sinh5t, sinh3t, cosht, sinh 4 tcosht, and sinht, we have

D ( t ) = 1 16 t ( sinh 5 t 5 sinh 3 t + 10 sinh t ) + 1 2 t ( sinh 3 t 3 sinh t ) + t 4 cosh t 1 16 ( cosh 5 t 3 cosh 3 t + 2 cosh t ) 1 4 t 3 ( sinh 3 t 3 sinh t ) 2 t 3 sinh t = 1 16 n = 0 5 2 n + 1 5 3 2 n + 1 + 10 ( 2 n + 2 ) ! t 2 n + 1 + 1 2 n = 0 3 2 n + 1 3 ( 2 n + 1 ) ! t 2 n + 2 + n = 0 1 ( 2 n ) ! t 2 n + 4 1 16 n = 0 5 2 n 3 3 2 n + 2 ( 2 n ) ! t 2 n 1 4 n = 0 3 2 n + 1 3 ( 2 n + 1 ) ! t 2 n + 4 2 n = 0 1 ( 2 n + 1 ) ! t 2 n + 4 = 1 16 n = 3 l n ( 2 n + 4 ) ! t 2 n + 4 ,

where

l n = ( 2 n + 4 ) ( 5 2 n + 3 5 3 2 n + 3 + 10 ) + 8 ( 2 n + 4 ) ( 3 2 n + 3 3 ) + 16 ( 2 n + 4 ) ( 2 n + 3 ) ( 2 n + 2 ) ( 2 n + 1 ) ( 5 2 n + 4 3 3 2 n + 4 + 2 ) 4 ( 2 n + 4 ) ( 2 n + 3 ) ( 2 n + 2 ) ( 3 2 n + 1 3 ) 32 ( 2 n + 4 ) ( 2 n + 3 ) ( 2 n + 2 ) = ( 250 n 125 ) 25 n + ( 279 462 n 432 n 2 96 n 3 ) 9 n + 256 n 4 + 1 , 120 n 3 + 1 , 520 n 2 + 532 n 154 , n = 3 , 4 , .

Using a basic differential method, we can easily prove

f ( x ) ( 250 x 125 ) 25 x + ( 279 462 x 432 x 2 96 x 3 ) 9 x + 256 x 4 + 1 , 120 x 3 + 1 , 520 x 2 + 532 x 154 > 0

on [3,). This leads to l n >0 for n=3,4, , and D(t)>0. So, the proof of Lemma 2.2 is complete. □

3 Proof of Theorem 1.3

Let

F(t) ( t sinh t e t coth t 1 ) p 1 ( t coth t ) p 1 = f 1 ( t ) f 1 ( 0 + ) g 1 ( t ) g 1 ( 0 + ) ,

where f 1 (t)= ( t sinh t e t coth t 1 ) p and g 1 (t)= ( t coth t ) p . Then

k 1 (t) f 1 ( t ) g 1 ( t ) = e p ( t coth t 1 ) ( cosh t ) p 1 sinh 2 t t 2 sinh t ( sinh t cosh t t ) .

We compute

k 1 (t)= e p ( t coth t 1 ) ( cosh t ) p u 1 ( t ) ( sinh t ) 3 ( sinh t cosh t t ) 2 ,

where

u 1 ( t ) = 2 t 2 sinh 4 t cosh t + sinh 4 t cosh t 4 t sinh 5 t 3 t sinh 3 t + 3 t 2 sinh 2 t cosh t t 3 sinh t p ( t sinh 5 t + 2 t sinh 3 t + t 4 cosh t sinh 4 t cosh t t 3 sinh 3 t 2 t 3 sinh t ) = 2 t 2 sinh 4 t cosh t + sinh 4 t cosh t 4 t sinh 5 t 3 t sinh 3 t + 3 t 2 sinh 2 t cosh t t 3 sinh t p D ( t ) .

If 0<p8/5, by Lemma 2.2 we have

5 u 1 ( t ) 10 t 2 sinh 4 t cosh t + 13 sinh 4 t cosh t 28 t sinh 5 t 46 t sinh 3 t + 30 t 2 sinh 2 t cosh t + 6 t 3 sinh t 8 t 4 cosh t + 8 t 3 sinh 3 t = n = 3 h n 16 ( 2 n + 4 ) ! t 2 n + 4 ,

where

h n = 10 ( 2 n + 4 ) ( 2 n + 3 ) ( 5 2 n + 2 3 3 2 n + 2 + 2 ) + 13 ( 5 2 n + 4 3 3 2 n + 4 + 2 ) 28 ( 2 n + 4 ) ( 5 2 n + 3 5 3 2 n + 3 + 10 ) 184 ( 2 n + 4 ) ( 3 2 n + 3 3 ) + 120 ( 2 n + 4 ) ( 2 n + 3 ) ( 3 2 n + 2 1 ) + 96 ( 2 n + 4 ) ( 2 n + 3 ) ( 2 n + 2 ) ( 2 n + 1 ) 2 n 128 ( 2 n + 4 ) ( 2 n + 3 ) ( 2 n + 2 ) ( 2 n + 1 ) + 96 ( 2 n + 4 ) ( 2 n + 3 ) ( 2 n + 2 ) ( 3 2 n 1 ) = ( 1 , 000 n 2 3 , 500 n 2 , 875 ) 25 n + ( 768 n 3 + 6 , 696 n 2 + 13 , 956 n + 4 , 113 ) 9 n + ( 2 n + 4 ) ( 2 n + 3 ) ( 2 n + 2 ) ( 2 n + 1 ) ( 192 n 128 ) 96 ( 2 n + 4 ) ( 2 n + 3 ) ( 2 n + 2 ) 100 ( 2 n + 4 ) ( 2 n + 3 ) + 272 ( 2 n + 4 ) + 26 > 0

for n=3,4, .

We have u 1 (t)>0 for 0<p8/5. So, k 1 (t)>0 for t>0, and f 1 (t)/ g 1 (t)= k 1 (t) is increasing on (0,+). Hence, F(t) is increasing on (0,+) by Lemma 2.1. At the same time, lim t 0 + F(t)=1/2 and lim t + F(t)= ( 2 / e ) p . So, the proof of Theorem 1.3 is complete.

4 Proof of Theorem 1.4

Let

S(t) ( sinh t t e 1 t coth t ) p 1 e p ( 1 t coth t ) 1 = f 2 ( t ) f 2 ( 0 + ) g 2 ( t ) g 2 ( 0 + ) ,

where f 2 (t)= ( sinh t t e 1 t coth t ) p and g 2 (t)= e p ( 1 t coth t ) . Then

k 2 (t) f 2 ( t ) g 2 ( t ) = ( sinh t t ) p 1 ( sinh t ) 3 t 2 sinh t t 2 ( sinh t cosh t t ) ,

and

k 2 (t)= ( sinh t t ) p 2 u 2 ( t ) t 4 ( sinh t cosh t t ) 2 ,

where

where e n =1( d n / c n ) and

Let

Then

e n =1 d n c n = j ( n ) i ( n ) .

Let Δ(n)=286i(n)693j(n). Then

Δ ( n ) = ( 741 , 313 n 5 , 759 , 424 ) 36 n + 16 n [ 2 , 275 , 328 ( 2 n + 5 ) + 4 , 009 , 984 + 70 , 400 ( 2 n + 5 ) ( 2 n + 4 ) ( 2 n + 3 ) 532 , 224 ( 2 n + 5 ) ( 2 n + 4 ) ] + 4 n [ 9 , 152 ( 2 n + 5 ) ( 2 n + 4 ) ( 2 n + 3 ) ( 2 n + 2 ) 62 , 656 ( 2 n + 5 ) ( 2 n + 4 ) ( 2 n + 3 ) + 133 , 056 ( 2 n + 5 ) ( 2 n + 4 ) 610 , 016 ( 2 n + 5 ) 156 , 640 ] .

First, we check that Δ(n)>0 for n=3,4,5,6,7; second, we can easily obtain that Δ(n)>0 for n8. So, we have that Δ(n)>0 for n=3,4, .

So, we have u 2 (t)>0 for p286/693. So, k 2 (t)>0 for t>0, and f 2 (t)/ g 2 (t)= k 2 (t) is increasing on (0,+). Hence, S(t) is increasing on (0,+) by Lemma 2.1 when p286/693. At the same time, lim t 0 + S(t)=1/2 and lim t + S(t)=1. So, the proof of Theorem 1.4 is complete.

5 Applications of theorems

In this section, we assume that x and y are two different positive numbers. Let A(x,y), G(x,y), L(x,y), and I(x,y) be the arithmetic, geometric, logarithmic, and identric means, respectively. Without loss of generality, we set 0<x<y. By the transformation t=(log(y/x))/2, we can compute and obtain

where t>0.

Now, the four results in Section 1 are equivalent to the following ones for four classical means.

Theorem 5.1 Let p4/5, and x and y be positive real numbers with xy. Then

α A p (x,y)+(1α) G p (x,y)< L p (x,y)<β A p (x,y)+(1β) G p (x,y)
(5.1)

holds if and only if α0 and β1/3.

Theorem 5.1 can deduce the following one, which is from Zhu [8].

Corollary 5.2 ([[8], Theorem 1])

Let p1, and x and y be positive real numbers with xy. Then

α A p (x,y)+(1α) G p (x,y)< L p (x,y)<β A p (x,y)+(1β) G p (x,y)
(5.2)

holds if and only if α0 and β1/3.

When letting p=1 in Theorem 5.1, one can obtain the result (see [1214], [[15], Theorem 1]).

Corollary 5.3 Let x and y be positive real numbers with xy. Then

αA(x,y)+(1α)G(x,y)<L(x,y)<βA(x,y)+(1β)G(x,y)
(5.3)

holds if and only if α0 and β1/3.

When letting β=1/3 in the right-hand inequality of (5.3), one can obtain the well-known inequality by Carlson [16]

L(x,y)< 1 3 A(x,y)+ 2 3 G(x,y).
(5.4)

Theorem 5.4 Let p>0. Then

(1) if 0<p6/5, the double inequality

α A p (x,y)+(1α) G p (x,y)< I p (x,y)<β A p (x,y)+(1β) G p (x,y)
(5.5)

holds if and only if α2/3 and β ( 2 / e ) p ;

(2) if p2, the double inequality

α A p (x,y)+(1α) G p (x,y)< I p (x,y)<β A p (x,y)+(1β) G p (x,y)
(5.6)

holds if and only if α ( 2 / e ) p and β2/3.

The part (2) of Theorem 5.4 is a result of Trif [17].

When letting p=2 and β=2/3 in the right-hand inequality of (5.6), one can obtain the following result, which is from Sándor and Trif [18].

I 2 (x,y)< 2 3 A 2 (x,y)+ 1 3 G 2 (x,y).
(5.7)

When letting p=1 in the double inequality (5.5), one can obtain the following result (see [12], [[15], Theorem 2]).

Corollary 5.5 Let x and y be positive real numbers with xy. Then

αA(x,y)+(1α)G(x,y)<I(x,y)<βA(x,y)+(1β)G(x,y)
(5.8)

holds if and only if α2/3 and β2/e.

When letting α=2/3 in the left-hand inequality in (5.8), one can obtain the following result, which is from Sándor [19].

2 3 A(x,y)+ 1 3 G(x,y)<I(x,y).
(5.9)

Theorem 5.6 Let 0<p8/5, x and y be positive real numbers with xy. Then

α A p (x,y)+(1α) L p (x,y)< I p (x,y)<β A p (x,y)+(1β) L p (x,y)
(5.10)

holds if and only if α1/2 and β ( 2 / e ) p .

Theorem 5.6 can deduce the following result (see Zhu [15]).

Corollary 5.7 ([[15], Theorem 3])

Let x and y be positive real numbers with xy. Then

αA(x,y)+(1α)L(x,y)<I(x,y)<βA(x,y)+(1β)L(x,y)
(5.11)

holds if and only if α1/2 and β2/e.

When letting α=1/2 in the left-hand inequality of (5.11), one can obtain the following result, which is from Sándor [4, 19].

I(x,y)> A ( x , y ) + L ( x , y ) 2 .
(5.12)

Finally, we give the bounds for L p (x,y) in terms of G p (x,y) and I p (x,y), and obtain the following new result.

Theorem 5.8 Let x and y be positive real numbers with xy, and p286/693. Then

α G p (x,y)+(1α) I p (x,y)< L p (x,y)<β G p (x,y)+(1β) I p (x,y)
(5.13)

holds if and only if β1/2 and α1.

Theorem 5.8 can deduce a result of Zhu [15]:

Corollary 5.9 ([[15], Theorem 4])

Let x and y be positive real numbers with xy. Then

αG(x,y)+(1α)I(x,y)<L(x,y)<βG(x,y)+(1β)I(x,y)
(5.14)

holds if and only if β1/2 and α1.

Obviously, the right-hand side of (5.14) is an extension of the following inequality:

L(x,y)< 1 2 ( G ( x , y ) + I ( x , y ) ) ,
(5.15)

which was given by Alzer [5].

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Zhu, L. New inequalities for hyperbolic functions and their applications. J Inequal Appl 2012, 303 (2012). https://doi.org/10.1186/1029-242X-2012-303

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Keywords

  • hyperbolic sine
  • hyperbolic cosine
  • hyperbolic cotangent
  • geometric mean
  • logarithmic mean
  • identric mean
  • arithmetic mean
  • best constants