ON GENERALIZATIONS AND REFINEMENTS OF JORDAN TYPE INEQUALITY

SHAN-HE WU

Abstract. In this paper, we give some refinements and generalizations of Jordan’s inequality, we also improve some known results.

1. Introduction and Lemmas

The following inequality is well-known Assume 0 < | x | ≤ \frac{π}{2} , then

\frac{2}{π} ≤ \frac{sin x}{x} < 1.  (1)

It is called Jordan’s inequality (see [?]). Jordan’s inequality is an important inequality in calculus and trigonometry, It has broad applications in theory of limit, There are many generalizations and refinements that had been published in recent years (see [?] [?] [?] [?]).

In this paper, we give some new generalizations and refinements of Jordan’s Inequality. Firstly, we prove following Lemmas.

Lemma 1. Assume 0 < β < α, then the function φ(x) = \frac{sin αx}{sin βx} is decreasing on interval (0, \frac{π}{α}), and ϕ(x) = \frac{tan αx}{tan βx} is increasing on interval (0, \frac{π}{2α}).

Proof. We note that φ is a differentiable function with

φ′(x) = \frac{α cos αx sin βx − β cos βx sin αx}{sin^{2} βx}, x ∈ (0, \frac{π}{α}).

Let g(x) = α cos αx sin βx − β cos βx sin αx, then g′(x) = (β^{2} − α^{2}) sin αx sin βx. Since 0 < β < α, 0 < x < \frac{π}{α}, we have g′(x) < 0, x ∈ (0, \frac{π}{α}). Consequently, g is decreasing on interval (0, \frac{π}{α}), and then we have g(x) < g(0) = 0 for any x ∈ (0, \frac{π}{α}), we deduce that φ′(x) < 0 for any x ∈ (0, \frac{π}{α}), it shows that φ is decreasing on

interval (0, \frac{π}{α}).

We note that ϕ is also a differentiable function with

ϕ′(x) = \frac{1}{2} sec^{2} αx csc^{2} βx(α sin 2βx − β sin 2αx), x ∈ (0, \frac{π}{2α}).

Let u(x) = α sin 2βx − β sin 2αx, then u′(x) = 2αβ (cos 2βx − cos 2αx). Since 0 < β < α, 0 < x < \frac{π}{2α}, we have u′(x) > 0, x ∈ (0, \frac{π}{2α}).

2000 Mathematics Subject Classification. 26D15, 26D10, 39B62, 52A40.

Key words and phrases. Jordan’s inequality, refinement, generalization, application, convex function, concave function.

This paper was typeset using AMS-L^{A}TEX.

Consequently, u is increasing on interval (0, \frac{π}{2α}), and then we have

u(x) > u(0) = 0 for any x ∈ (0, \frac{π}{2α}),

we deduce that ϕ′(x) > 0 for any x ∈ (0, \frac{π}{2α}), it shows that ϕ is increasing on interval (0, \frac{π}{2α}).

The proof is complete. □

Lemma 2. Assume 0 < β < α, 0 < |αx| ≤ \frac{π}{2} , then

2 | \frac{x}{π} | ≤ | (sin βx)/(α sin \frac{πβ}{2α}) | ≤ | \frac{sin αx}{α} | < | \frac{sin βx}{β} | < | x | .  (2)

Proof. Case 1, when 0 < x ≤ \frac{π}{2α}. By Lemma ??, we see that φ(x) = \frac{sin αx}{sin βx} is decreasing on interval (0, \frac{π}{2α}). Since

lim_{x→0^{+}} \frac{sin αx}{sin βx} = lim_{x→0^{+}} \frac{α}{β} · \frac{sin αx}{αx} · \frac{βx}{sin βx} = \frac{α}{β}, φ( \frac{π}{2α} ) = csc \frac{πβ}{2α} ,

thus

\frac{α}{β} > \frac{sin αx}{sin βx} ≥ csc \frac{πβ}{2α}  (3)

for any 0 < β < α and 0 < x ≤ \frac{π}{2α}.

Since 0 < x ≤ \frac{π}{2α}, 0 < β < π/(2(\frac{π}{2α})), by inequality (??), we obtain

(\frac{π}{2α})/x ≥ (sin \frac{π}{2α}β)/(sin xβ),

that is

\frac{2x}{π} ≤ (sin βx)/(α sin \frac{πβ}{2α}).  (4)

Using Jordan’s inequality, we have

\frac{sin βx}{β} < x.  (5)

Combining inequalities (??),(??),(??), we get

0 < \frac{2x}{π} ≤ (sin βx)/(α sin \frac{πβ}{2α}) ≤ \frac{sin αx}{α} < \frac{sin βx}{β} < x,  (6)

that is

2 | \frac{x}{π} | ≤ | (sin βx)/(α sin \frac{πβ}{2α}) | ≤ | \frac{sin αx}{α} | < | \frac{sin βx}{β} | < | x | .

Case 2, when − \frac{π}{2α} ≤ x < 0, then 0 < − x ≤ \frac{π}{2α}. By the result of Case 1, we have

2 | \frac{−x}{π} | ≤ | (sin(−βx))/(α sin \frac{πβ}{2α}) | ≤ | \frac{sin(− αx)}{α} | < | \frac{sin(− βx)}{β} | < | −x | .

that is

2 | \frac{x}{π} | ≤ | (sin βx)/(α sin \frac{πβ}{2α}) | ≤ | \frac{sin αx}{α} | < | \frac{sin βx}{β} | < | x | .

The proof is complete. □

Lemma 3. The function f(x) = ln \frac{sin x}{x} is concave on interval (0, \frac{π}{2} ), and g(x) = ln \frac{tan x}{x} is convex on interval (0, \frac{π}{2} ).

Proof. f is a second order differentiable function with

f′′(x) = \frac{csc^{2} x}{x^{2}}( sin^{2} x − x^{2}), x ∈ (0, \frac{π}{2} ).

Using Jordan’s inequality, we get f′′(x) < 0 for any x ∈ (0, \frac{π}{2} ). Therefore f is concave on interval (0, \frac{π}{2} ).

By g is also a second order differentiable function with

g′′(x) = \frac{csc^{2} 2x}{x^{2}}(sin^{2} 2x − 4x^{2} cos 2x), x ∈ (0, \frac{π}{2} ).

Firstly, suppose \frac{π}{4} ≤ x < \frac{π}{2} , it is easy to see that g′′(x) > 0. Secondly, consider the case of 0 < x < \frac{π}{4} . Now

g′′(x) = \frac{csc^{2} 2x}{x^{2}}(sin^{2} 2x − 4x^{2} cos 2x) = \frac{csc^{2} 2x cos 2x}{x^{2}}(sin 2x tan 2x − 4x^{2}).

Let q(x) = sin 2x tan 2x − 4x^{2}, we have

q′(x) = 2 cos 2x tan 2x + 2 sin 2x sec^{2} 2x − 8x,

q′′(x) = 4 sec 2x(cos 2x − 1)^{2} + 8 sin 2x tan 2x sec^{2} 2x > 0, x ∈ (0, \frac{π}{4} ),

Hence q′(x) is increasing on interval (0, \frac{π}{4} ), and then we have q′(x) > q′(0) = 0 for any x ∈ (0, \frac{π}{4} ).

Thus q(x) is increasing on interval (0, \frac{π}{4} ), further we get q(x) > q(0) = 0 for any x ∈ (0, \frac{π}{4} ).

Therefore g′′(x) > 0 for any x ∈ (0, \frac{π}{4} ), actually g′′(x) > 0 for any x ∈ (0, \frac{π}{2} ). It shows that g is convex on interval (0, \frac{π}{2} ).

The proof is complete. □

2. Main Results and Their Proof

Theorem 1. Assume 0 < | y | < | x |, 0 < |λx| ≤ \frac{π}{2} , then the following inequality holds

2 | \frac{λ}{π} | ≤ | (sin λy)/(x sin \frac{πy}{2x}) | ≤ | \frac{sin λx}{x} | < | \frac{sin λy}{y} | < | λ | .  (7)

Proof. By hypothesis, we have

0 < | y | < | x | , 0 < | |x| λ | ≤ \frac{π}{2} ,

Using Lemma ??, we get

2 | \frac{λ}{π} | ≤ | (sin λ |y|)/(|x| sin \frac{π}{2} |\frac{y}{x}|) | ≤ | \frac{sin(λ | x |)}{| x |} | < | \frac{sin(λ | y |)}{| y |} | < | λ | .

By identical equation

| (sin λ |y|)/(|x| sin \frac{π}{2} |\frac{y}{x}|) | = | (sin λy)/(x sin \frac{πy}{2x}) | ,

| \frac{sin(λ | x |)}{| x |} | = | \frac{sin λx}{x} | , | \frac{sin(λ | y |)}{| y |} | = | \frac{sin λ y}{y} | ,

we immediately obtain inequality (??). The proof of Theorem ?? is complete. □ From Theorem ?? we get

Corollary 1. Assume 0 < | λx | ≤ \frac{π}{2} , then the following inequality holds

2 | \frac{λ}{π} | ≤ | \frac{sin λx}{x} | < | λ | .  (8)

Remark. If we choose λ = 1 in Corollary ??, inequality (??) become Jordan’s inequality. Corollary 2. Assume 0 < | y | < | x | ≤ \frac{π}{2} , then the following inequality holds

\frac{2}{π} ≤ (sin y)/(x sin \frac{πy}{2x}) ≤ \frac{sin x}{x} < \frac{sin y}{y} < 1.  (9)

Proof. Let λ = 1 in Theorem ??, we have

\frac{2}{π} ≤ | (sin y)/(x sin \frac{πy}{2x}) | ≤ | \frac{sin x}{x} | < | \frac{sin y}{y} | < 1  (10)

for any 0 < | y | < | x | ≤ \frac{π}{2} ,

By inequality (??) and since

\frac{sin x}{x} > 0, \frac{sin y}{y} > 0, (sin y)/(x sin \frac{πy}{2x}) = \frac{π}{2} · \frac{sin y}{y} · ((\frac{πy}{2x})/(sin \frac{πy}{2x})) > 0, we obtain inequality (??) □

Remark. The inequality (??) is a refinement of Jordan’s inequality . Theorem 2. Assume 0 < β < α, 0 < |αx| < \frac{π}{2} , then the following inequality holds

| tan αx | > \frac{α}{β} | tan βx | > α | x | > \frac{α}{β} | sin βx | > | sin αx | > csc \frac{πβ}{2α} | sin βx | . (11)

Proof. From Lemma ??, we get

\frac{α}{β} | sin βx | > | sin αx | > csc \frac{πβ}{2α} | sin βx | .  (12)

Firstly, we prove following inequality

| tan αx | > \frac{α}{β} | tan βx |  (13)

for any 0 < | αx | < \frac{π}{2} .

Case 1, when 0 < x ≤ \frac{π}{2α}. By Lemma ??, we see that ϕ(x) = \frac{tan αx}{tan βx} is increasing on interval (0, \frac{π}{2α}).

Since

lim_{x→0^{+}} \frac{tan αx}{tan βx} = lim_{x→0^{+}} \frac{α}{β} · \frac{sin αx}{αx} · \frac{βx}{sin βx} · \frac{cos βx}{cos αx} = \frac{α}{β},

therefore \frac{tan αx}{tan βx} > \frac{α}{β}, it yields that tan αx > \frac{α}{β} tan βx for any 0 < x < \frac{π}{2α}, By

tan αx = | tan αx | , tan βx = | tan βx | ,

we get inequality (??)

Case 2, when − \frac{π}{2α} < x < 0, then 0 < − x < \frac{π}{2α}. By the result of Case 1, we have

| tan(− αx) | > \frac{α}{β} | tan(− βx) | ,

which proves inequality (??).

By the well-known inequality (see [?])

| tan x | > | x | > | sin x |  (14)

for any 0 < | x | < \frac{π}{2} , we get

\frac{α}{β} | tan βx | > \frac{α}{β} | βx | = α | x | , \frac{α}{β} | sin βx | < \frac{α}{β} | βx | = α | x | ,

Combining inequalities (??),(??) and above inequalities, we immediately obtain (??). The proof of Theorem ?? is complete. □

Let α = 1 in Theorem ??, we obtain

Corollary 3. Assume 0 < β < 1, 0 < | x | < \frac{π}{2} , then the following inequality holds

| tan x | > \frac{1}{β} | tan βx | > | x | > \frac{1}{β} | sin βx | > | sin x | .  (15)

Remark. The inequality (??) is a refinement of inequality(??).

Corollary 4. Assume 0 < β < α, 0 < |αx| < \frac{π}{2} , then the following inequality holds

\frac{tan αx}{tan βx} > \frac{α}{β} > \frac{sin αx}{sin βx} > csc \frac{πβ}{2α} .  (16)

Proof. Case 1, when 0 < x < \frac{π}{2α}. Then

sin αx > 0, sin βx > 0, tan αx > 0, tan βx > 0,

From Theorem ??, we obtain inequality (??).

Case 2, when − \frac{π}{2α} < x < 0, then 0 < − x < \frac{π}{2α}, By the result of Case 1, we get

\frac{tan(− αx)}{tan(− βx)} > \frac{α}{β} > \frac{sin(− αx)}{sin(− βx)} > csc \frac{πβ}{2α} ,

that is inequality (??). □

Let x = 1 in Corollary ??, we obtain

Corollary 5. Assume 0 < β < α < \frac{π}{2} , then the following inequality holds

\frac{α}{β} > \frac{sin α}{sin β} > csc \frac{πβ}{2α} .  (17)

Remark. Note that csc \frac{πβ}{2α} = (sin \frac{πβ}{2α})^{−1} > (\frac{πβ}{2α})^{−1} = \frac{2α}{πβ} for any 0 < β < α < \frac{π}{2} . Clearly, inequality (??) sharpens following Garnir’s inequality(see [?])

\frac{α}{β} > \frac{sin α}{sin β} > \frac{2α}{πβ} .  (18)

Theorem 3. Assume 0 < |x_{i}| < \frac{π}{2} , i = 1, 2, ⋯, n, then the following inequality holds

|tan x_{1} tan x_{2} ⋯ tan x_{n}| ≥ |x_{1}x_{2} ⋯ x_{n}((tan \frac{|x_{1}| + ⋯ + |x_{n}|}{n})/(\frac{|x_{1}| + ⋯ + |x_{n}|}{n}))^{n}|

> |x_{1}x_{2} ⋯ x_{n}| > |x_{1}x_{2} ⋯ x_{n}( (sin \frac{|x_{1}| + ⋯ + |x_{n}|}{n})/(\frac{|x_{1}| + ⋯ + |x_{n}|}{n}) )^{n}|

≥ |sin x_{1} sin x_{2} ⋯ sin x_{n}| .  (19)

Proof. By Lemma ?? and Jensen’s inequality(see [?]), we have

\frac{1}{n}( ln \begin{matrix}tan & |x_{1}| \\ | & x_{1} & | \end{matrix} + ⋯ + ln \begin{matrix}tan & |x_{n}| \\ | & x_{n} & | \end{matrix}) ≥ ln((tan \frac{|x_{1}| + ⋯ + |x_{n}|}{n})/(\frac{|x_{1}| + ⋯ + |x_{n}|}{n})),

\frac{1}{n}( ln \begin{matrix}sin & |x_{1}| \\ | & x_{1} & | \end{matrix} + ⋯ + ln \begin{matrix}sin & |x_{n}| \\ | & x_{n} & | \end{matrix}) ≤ ln((sin \frac{|x_{1}| + ⋯ + |x_{n}|}{n})/(\frac{|x_{1}| + ⋯ + |x_{n}|}{n})),

It is obvious that above inequalities are equivalent to following inequalities

tan |x_{1}| ⋯ tan |x_{n}| ≥ |x_{1}x_{2} ⋯ x_{n}( (tan \frac{|x_{1}| + ⋯ + |x_{n}|}{n})/(\frac{|x_{1}| + ⋯ + |x_{n}|}{n}) )^{n}| ,

sin |x_{1}| ⋯ sin |x_{n}| ≤ |x_{1}x_{2} ⋯ x_{n}( (sin \frac{|x_{1}| + ⋯ + |x_{n}|}{n})/(\frac{|x_{1}| + ⋯ + |x_{n}|}{n}) )^{n}| .

Now

|tan x_{1} tan x_{2} ⋯ tan x_{n}| = tan |x_{1}| tan |x_{2}| ⋯ tan |x_{n}| ,

|sin x_{1} sin x_{2} ⋯ sin x_{n}| = sin |x_{1}| sin |x_{2}| ⋯ sin |x_{n}| ,

|( (tan \frac{|x_{1}| + ⋯ + |x_{n}|}{n})/(\frac{|x_{1}| + ⋯ + |x_{n}|}{n}) )^{n}| > 1,

|( (sin \frac{|x_{1}| + ⋯ + |x_{n}|}{n})/(\frac{|x_{1}| + ⋯ + |x_{n}|}{n}) )^{n}| < 1.

Base on the above results, we immediately obtain inequality (??). The proof of Theorem ?? is complete. □

Remark. Inequality (??) involves n variables x_{1}, x_{2}, ⋯, x_{n}. Clearly, it is the refinement and generalization of Jordan’s inequality.

References

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[2] F.- Qi, Extensions and sharpening of Jordan’s and Kober’s Inequality. Journal of Mathematics For Technology. 12 (1996), No. 4, 98–102. (Chinese).

[3] F.- Qi and Q.-D. Hao, Refinements and Extensions and sharpening of Jordan’s and Kober’s Inequality. Math. Inform. Quart. 8(1998), No. 3, 116–120.

[4] F. Yuefeng, Jordan’s Inequality. Math. Mag. 69(1996), 126.

[5] F.- Qi, Ch.-P. Chen and J.-W. Zhao, Three inequalities involving hyperbolically trigonometric functions. RGMIA Res. Rep. Coll. 5(2002), Art. 2.

[6] J.-Ch. Kuang, Applied Inequalities, 2nd ed., Hunan Education, Changsha, China, 1993. 309–310. (Chinese).

[7] D. S. Mitrinović, J. E. Pečarić and A. M. Fink, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993.

(SH.-H. Wu) Department of Mathematics, Longyan College, Longyan Fujian 364012, CHINA

   E-mail address: wushanhe@yahoo.com.cn
   URL: http://rgmia.vu.edu.au/members/Wu2.htm