INEQUALITIES FOR WEIERSTRASS PRODUCTS

Toufik Mansour ^{1}

Department of Mathematics, Chalmers University of Technology, S-41296 Göteborg, Sweden

                              toufik@math.chalmers.se

Abstract

The aim of this note is to find new inequalities for the Weierstrass products ∏_{i=1}^{n}(1+x_{i}) and ∏_{i=1}^{n}(1−x_{i}),

which are an analogues of Alzer’s inequalities [A].

1. Introduction

Several authors interested in Weierstrass products ∏_{i=1}^{n}(1 + x_{i}) and ∏_{i=1}^{n}(1 − x_{i}). In [B, Pages 104–105]

proved

1 − ∑_{i=1}^{n} x_{i} ≤ ∏_{i=1}^{n}(1 − x_{i}) ≤ (1 + ∑_{i=1}^{n} x_{i})^{−1} ,

and

1 + ∑_{i=1}^{n} x_{i} ≤ ∏_{i=1}^{n}(1 + x_{i}) ≤ (1 − ∑_{i=1}^{n} x_{i})^{−1} ,

where x_{i} ∈ [0, 1], i = 1, 2, …, n, and in the second inequality assumed that ∑_{i=1}^{n} x_{i} < 1. In the last years these products attacked by many other inequalities which have been published. For example, Klamkin and Newman [KN] discovered several extensions and new inequalities:

(n + 1)^{n} ∏_{i=1}^{n} x_{i} ≤ ∏_{i=1}^{n}(1 + x_{i}), (n − 1)^{n} ∏_{i=1}^{n} x_{i} ≤ ∏_{i=1}^{n}(1 − x_{i}),

and in [K] presented another inequality:

(n + 1)^{n} ∏_{i=1}^{n}(1 − x_{i}) ≤ (n − 1)^{n} ∏_{i=1}^{n}(1 + x_{i}),

where x_{i} ∈ [0, 1], i = 1, 2, …, n, and ∑_{i=1}^{n} x_{i} = 1. An another example, the Ky Fan’s (see [B, Page 5]) inequality:

(1.1)  (∏_{i=1}^{n} \frac{x_{i}}{1 − x_{i}} )^{1/n} ≤ \frac{∑_{i=1}^{n} x_{i}}{∑_{i=1}^{n}(1 − x_{i})},

1Research financed by EC’s IHRP Programme, within the Research Training Network ”Algebraic Combinatorics in Europe”, grant HPRN-CT-2001-00272

where x_{i} ∈ (0, \frac{1}{2}], i = 1, 2, …, n. In this paper, we are interesting in the Alzer’s inequalities [A] of Weierstrass Products. Here, we present a generalization for the Alzer’s inequalities by using the same arguments proof as in Alzer’s paper [A]. In Section 2 we establish new inequalities for Weierstrass products which are an analogues for these inequalities in [A]. Finally, we apply the applications of [A] to get new inequalities for trigonometric and integral sums and products.

2. Inequalities for Weierstrass products

Let us start by proving two lemmas which are an analogues of the two lemmas in [A]. Lemma 2.1. For all real x_{i} ∈ [0, 1], i = 1, 2, …, n, and a ∈ [0, 1], we have

(2.1)  ∏_{i=1}^{n} x_{i} [1 + a ∑_{i=1}^{n}(1 − x_{i})] + ∏_{i=1}^{n}(1 − x_{i}) [1 + a ∑_{i=1}^{n} x_{i}] ≤ \frac{1}{2}a + 1.

Proof. Using the same arguments as in the proof of [A, Lemma 1] we get as follows. Let

S(x_{1}, …, x_{n}) = ∏_{i=1}^{n} x_{i} [1 + a ∑_{i=1}^{n}(1 − x_{i})] + ∏_{i=1}^{n}(1 − x_{i}) [1 + a ∑_{i=1}^{n} x_{i}] ,

since a, x_{i} ∈ [0, 1] for all i, we arrive at

\begin{matrix}S(x_{1}, …, x_{n}) − S(x_{1}, …, x_{n−1}) = −a(1 − x_{n}) ∏_{i=1}^{n−1} x_{i} ∑_{i=1}^{n}(1 − x_{i}) − (1 − a)(1 − x_{n}) ∏_{i=1}^{n−1} x_{i} \\ −ax_{n} ∏_{i=1}^{n−1}(1 − x_{i}) ∑_{i=1}^{n} x_{i} − (1 − a)x_{n} ∏_{i=1}^{n−1}(1 − x_{i}) ≤ 0. \end{matrix}

Hence, inductively on n we have

S(x_{1}, …, x_{n}) ≤ S(x_{1}) = −2a (x_{1} − \frac{1}{2})^{2} + 1 + \frac{1}{2}a ≤ \frac{1}{2}a + 1,

as requested. □

The next lemma is a counterpart of Ky Fan’s Inequality 1.1 and analogues of [A, Lemma 2]. Lemma 2.2. For all real x_{i} ∈ (0, \frac{1}{2}], i = 1, 2, …, n, and a ≥ 0, we have

(2.2)  \frac{∏_{i=1}^{n} x_{i}}{∏_{i=1}^{n}(1 − x_{i})} ≤ \frac{1 + a ∑_{i=1}^{n} x_{i}}{1 + a ∑_{i=1}^{n}(1 − x_{i})},

with equality if and only if x_{i} = \frac{1}{2} for all i = 1, 2, …, n.

Proof. To this theorem we present two different proofs: by induction argument, and by Ky Fan’s Inequality 1.1.

(1) It is easy to see that the equality holds in 2.2 for x_{i} = \frac{1}{2} where i = 1, 2, …, n. Now, we prove the lemma by induction on n. For n = 1, Inequality 2.2 yields

\frac{x_{1}}{1 − x_{1}} ≤ \frac{1 + ax_{1}}{1 + a(1 − x_{1})},

equivalently, x_{1} ≤ \frac{1}{2} and the sign of equality holds if and only if x_{1} = \frac{1}{2}. Now, let us suppose that

0 < x_{1} ≤ x_{2} ≤ ⋯ ≤ x_{n} ≤ x_{n+1} ≤ \frac{1}{2}, x_{1} < x_{n+1},

and further we assume that Inequality 2.2 is true for n. Therefore, by induction hypothesis we get

\frac{∏_{i=1}^{n+1} x_{i}}{∏_{i=1}^{n+1}(1 − x_{i})} ≤ \frac{1 + a ∑_{i=1}^{n} x_{i}}{1 + a ∑_{i=1}^{n}(1 − x_{i})} · \frac{x_{n+1}}{1 − x_{n+1}} .

If putting p = a ∑_{i=1}^{n} x_{i}, q = a ∑_{i=1}^{n}(1 − x_{i}), and x = x_{n+1}, then we have to prove that

\frac{x(1 + p)}{(1 − x)(1 + q)} < \frac{1 + p + ax}{1 + q + a(1 − x)},

equivalently,

a(q − p)x^{2} + x(2(1 + p)(1 + q) − a(q − p)) − (1 + p)(1 + q) < 0.

Since, p = q if and only if x_{1} = x_{2} = … = x_{n} = \frac{1}{2} we have p < q and since 0 ≤ x ≤ \frac{1}{2} we obtain

a(q − p)x^{2} + x(2(1 + p)(1 + q) − a(q − p)) − (1 + p)(1 + q) < −\frac{a(q − p)}{4} < 0.

Thus we have the desired result.

(2) Since x_{i} ≤ \frac{1}{2} for all i, we have that \frac{x_{i}}{1−x_{i}} ≤ 1. Therefore, by Inequality 1.1 we have

∏_{i=1}^{n} \frac{x_{i}}{1 − x_{i}} ≤ [∏_{i=1}^{n} \frac{x_{i}}{1 − x_{i}} ]^{1/n} ≤ \frac{∑_{i=1}^{n} x_{i}}{∑_{i=1}^{n}(1 − x_{i})} = \frac{a ∑_{i=1}^{n} x_{i}}{a ∑_{i=1}^{n}(1 − x_{i})}.

Hence, because of \frac{x}{y} ≤ 1 if and only if \frac{x}{y} ≤ \frac{1+x}{1+y} where x, y > 0, and since a ∑_{i=1}^{n} x_{i} ≤ a ∑_{i=1}^{n}(1 − x_{i}), we arrive at

∏_{i=1}^{n} \frac{x_{i}}{1 − x_{i}} ≤ \frac{1 + a ∑_{i=1}^{n} x_{i}}{1 + a ∑_{i=1}^{n}(1 − x_{i})}.

□

For example, Theorem 2.2, for a = 2, yields [A, Lemma 2]. An another example, Theorem 2.2, for a → ∞, yields

\frac{∏_{i=1}^{n} x_{i}}{∏_{i=1}^{n}(1 − x_{i})} ≤ \frac{∑_{i=1}^{n} x_{i}}{∑_{i=1}^{n}(1 − x_{i})},

Theorem 2.3. For all real x_{i} ∈ (0, \frac{1}{2}], i = 1, 2, …, n, (n ≥ 2), and 1 ≤ a ≤ \frac{3}{2}, we have

(2.3)  \frac{∏_{i=1}^{n}(1 + x_{i})}{∏_{i=1}^{n}(1 + (1 − x_{i}))} ≤ \frac{1 + a ∑_{i=1}^{n} x_{i}}{1 + a ∑_{i=1}^{n}(1 − x_{i})},

with equality if and only if x_{i} = \frac{1}{2} for all i = 1, 2, …, n.

Proof. It is easy to see that the equality holds in 2.3 for x_{i} = \frac{1}{2} where i = 1, 2, …, n. Therefore, it remains to prove for numbers x_{1}, x_{2}, …, x_{n} are not all equal to \frac{1}{2} then the Inequality 2.3 is valid with ”<” instead of ”≤”.

4 INEQUALITIES FOR WEIERSTRASS PRODUCTS

Let us prove Inequality 2.3 by induction on n. For n = 2 we have to prove, for 0 < x_{1} < x_{2} ≤ \frac{1}{2},

\frac{(1 + x_{1})(1 + x_{2})}{(2 − x_{1})(2 − x_{2})} < \frac{1 + ax_{1} + ax_{2}}{1 + 2a − ax_{1} − ax_{2}} ,

equivalently,

(3 − 2a + a(x_{1} + x_{2} − 2x_{1}x_{2}))(x_{1} + x_{2} − 1),

which is absolutely true for 0 < x_{1} < x_{2} ≤ \frac{1}{2} and 1 < a ≤ \frac{3}{2}. Now, let us suppose that

0 < x_{1} ≤ x_{2} ≤ ⋯ ≤ x_{n} ≤ x_{n+1} ≤ \frac{1}{2}, x_{1} < x_{n+1},

and further we suppose that Inequality 2.3 is hold for n. Therefore, by induction hypothesis we get

∏_{i=1}^{n+1} \frac{1 + x_{i}}{(1 + (1 − x_{i}))} ≤ \frac{1 + a ∑_{i=1}^{n} x_{i}}{1 + a ∑_{i=1}^{n}(1 − x_{i})} · \frac{1 + x_{n+1}}{2 − x_{n+1}} .

If putting p = a ∑_{i=1}^{n} x_{i}, q = a ∑_{i=1}^{n}(1 − x_{i}), and x = x_{n+1}, then we have to prove that

\frac{(1 + x)(1 + p)}{(2 − x)(1 + q)} < \frac{1 + p + ax}{1 + q + a(1 − x)},

equivalently,

a(q − p)x^{2} + 2x(1 + q)(1 + p − a) + (1 + p)(a − 1 − q) < 0.

Since, a = b if and only if x_{i} = \frac{1}{2} for i = 1, 2, …, n, we have p < q. Therefore, since x ≤ \frac{1}{2} and 1 < a ≤ \frac{3}{2} we obtain

a(q − p)x^{2} + 2x(1 + q)(1 + p − a) + (1 + p)(a − 1 − q) ≤ −\frac{3a(q − p)}{4} < 0,

which this completes the proof. □ For example, Theorem 2.3, for a = 1 yields the second inequality in [A, Theorem, Equation 2.2]. Corollary 2.4. For all real x_{i} ∈ [\frac{1}{2}, 1), i = 1, 2, …, n, (n ≥ 2), and 1 ≤ a ≤ \frac{3}{2}, we have

(2.4)  \frac{∏_{i=1}^{n}(1 + x_{i})}{∏_{i=1}^{n}(1 + (1 − x_{i}))} ≥ \frac{1 + a ∑_{i=1}^{n} x_{i}}{1 + a ∑_{i=1}^{n}(1 − x_{i})},

with equality if and only if x_{i} = \frac{1}{2} for all i = 1, 2, …, n. Proof. If we set x_{i} = 1−y_{i}, y_{i} ∈ [\frac{1}{2}, 1), i = 1, 2, …, n, in Inequality 2.3 then we arrive to Inequality 2.4.

□

Theorem 2.5. For all real x_{i} ∈ (0, \frac{1}{2}], i = 1, 2, …, n, (n ≥ 2), and \frac{2}{3} < a ≤ 1, we have

(2.5)  \frac{1 + a ∑_{i=1}^{n} x_{i}}{1 + a ∑_{i=1}^{n}(1 − x_{i})} ≤ \frac{1 + ∏_{i=1}^{n} x_{i}}{1 + ∏_{i=1}^{n}(1 − x_{i})},

with equality if and only if x_{i} = \frac{1}{2} for all i = 1, 2, …, n.

Proof. It is easy to see that the equality holds in 2.5 for x_{i} = \frac{1}{2} where i = 1, 2, …, n. Therefore, it remains to prove for numbers x_{1}, x_{2}, …, x_{n} are not all equal to \frac{1}{2} then the Inequality 2.5 is valid with ”<” instead of ”≤”.

Let us prove Inequality 2.5 by induction on n. By simplify Inequality 2.5 we arrive to

(2.6)  an − 2a ∑_{i=1}^{n} x_{i} + (an + 1) ∏_{i=1}^{n} x_{i} − ∏_{i=1}^{n}(1 − x_{i}) − a ∑_{i=1}^{n} x_{i} (∏_{i=1}^{n} x_{i} + ∏_{i=1}^{n}(1 − x_{i})) ≥ 0.

For n = 2 we have to prove, for 0 < x_{1} < x_{2} ≤ \frac{1}{2},

h(x_{2}) = 2a − 2a(x_{1} + x_{2}) + (2a + 1)x_{1}x_{2} − (1 − x_{1})(1 − x_{2}) − a(x_{1} + x_{2})(x_{1}x_{2} + (1 − x_{1})(1 − x_{2})),

where h : (x_{1}, \frac{1}{2}] → ℝ. So, h′′(x_{2}) = 2x(1 − 2x_{1}) > 0, and h′(x_{2}) = 0 if and only if

x_{2} = \frac{(x_{1} − 1)^{2}}{1 − 2x_{1}} + \frac{a − 1}{2a(1 − 2x_{1})} ≥ \frac{(x_{1} − 1)^{2}}{1 − 2x_{1}} − \frac{1}{4(1 − 2x_{1})} = \frac{3}{4} − \frac{1}{2}x_{1} ≥ \frac{1}{2}.

Therefore, h(x_{2}) is strictly decreasing on (x_{1}, \frac{1}{2}] which gives

h(x_{2}) ≥ h (\frac{1}{2}) = (\frac{1}{2} − x_{1}) (\frac{3a}{2} − 1) > 0.

Now, let us suppose that

0 < x_{1} ≤ x_{2} ≤ ⋯ ≤ x_{n} ≤ x_{n+1} ≤ \frac{1}{2}, x_{1} < x_{n+1},

and further we suppose that Inequality 2.3 is hold for n. Let us define a function f : (0, \frac{1}{2}] → ℝ by

\begin{matrix}f(x) = a(n + 1) − 2a ∑_{i=1}^{n} x_{i} − 2ax + (an + a + 1)x ∏_{i=1}^{n} x_{i} − (1 − x) ∏_{i=1}^{n}(1 − x_{i}) \\ −a (x + ∑_{i=1}^{n} x_{i}) (x ∏_{i=1}^{n} x_{i} + (1 − x) ∏_{i=1}^{n}(1 − x_{i})) , \end{matrix}

so we have to prove f(x_{n+1}) > 0. This by prove that f is a convex function, f′ (\frac{1}{2}) < 0, and f (\frac{1}{2}) > 0. The function f is a convex function because f′′(x) = 2a (∏_{i=1}^{n}(1 − x_{i}) − ∏_{i=1}^{n} x_{i}) ≥ 0. By direct calculations together with Lemma 2.1 we arrive at

f′ (\frac{1}{2}) = ∏_{i=1}^{n} x_{i} (1 + a ∑_{i=1}^{n} x_{i}) + ∏_{i=1}^{n}(1 − x_{i}) (1 + a ∑_{i=1}^{n} x_{i}) − 2a ≤ 1 + \frac{a}{2} − 2a ≤ 1 − \frac{3a}{2} < 0.

To prove that f (\frac{1}{2}) > 0, let us write

f (\frac{1}{2}) = u(x_{1}, …, x_{n}) + v(x_{1}, …, x_{n}) + \frac{1}{2}(1 − a) (∏_{i=1}^{n}(1 − x_{i}) − ∏_{i=1}^{n} x_{i}) ,

where

u(x_{1}, …, x_{n})

= an − 2a ∑_{i=1}^{n} +(an + 1) ∏_{i=1}^{n} x_{i} − ∏_{i=1}^{n}(1 − x_{i}) − a ∑_{i=1}^{n} x_{i} ( ∏_{i=1}^{n} x_{i} + ∏_{i=1}^{n}(1 − x_{i})) ,

v(x_{1}, …, x_{n})

= \frac{a}{4} ∏_{i=1}^{n}(1 − x_{i}) (1 + 2 ∑_{i=1}^{n} x_{i}) − \frac{a}{4} ∏_{i=1}^{n} x_{i} (1 + 2 ∑_{i=1}^{n}(1 − x_{i})) .

Since \frac{2}{3} < a ≤ 1 we have \frac{1}{2}(1 − a) (∏_{i=1}^{n}(1 − x_{i}) − ∏_{i=1}^{n} x_{i}) ≥ 0, and by induction hypothesis we get u(x_{1}, …, x_{n}) ≥ 0. Lemma 2.2, for a = 2, yields v(x_{1}, …, x_{n}) > 0. Hence, f (\frac{1}{2}) > 0 as requested. □

For example, Theorem 2.5, for a = 1 yields the third inequality in [A, Theorem, Equation 2.2]. Corollary 2.6. For all real x_{i} ∈ [\frac{1}{2}, 1), i = 1, 2, …, n, (n ≥ 2), and 1 ≤ a ≤ \frac{3}{2}, we have

(2.7)  \frac{1 + a ∑_{i=1}^{n} x_{i}}{1 + a ∑_{i=1}^{n}(1 − x_{i})} ≥ \frac{1 + ∏_{i=1}^{n} x_{i}}{1 + ∏_{i=1}^{n}(1 − x_{i})},

with equality if and only if x_{i} = \frac{1}{2} for all i = 1, 2, …, n.

Proof. If we set x_{i} = 1−y_{i}, y_{i} ∈ [\frac{1}{2}, 1), i = 1, 2, …, n, in Inequality 2.5 then we arrive to Inequality 2.7.

□

3. Applications

In this section we present two applications of the pervious results. In [M] presented a great number of inequalities involving trigonometric functions. Here, we can be formulated another two inequalities which are the analogue of [A, Corollary 1]. Corollary 3.1.

(i) For all x_{i} ∈ (0, \frac{π}{4} ], i = 1, 2, …, n, (n ≥ 2), and 1 ≤ a ≤ \frac{3}{2}, we have

\frac{∏_{i=1}^{n}(1 + sin^{2}(x_{i}))}{∏_{i=1}^{n}(1 + cos^{2}(x_{i}))} ≤ \frac{1 + a ∑_{i=1}^{n} sin^{2}(x_{i})}{1 + a ∑_{i=1}^{n} cos^{2}(x_{i})},

(ii) For all x_{i} ∈ [\frac{π}{4} , \frac{π}{2} ), i = 1, 2, …, n, (n ≥ 2), and 1 ≤ a ≤ \frac{3}{2}, we have

\frac{∏_{i=1}^{n}(1 + sin^{2}(x_{i}))}{∏_{i=1}^{n}(1 + cos^{2}(x_{i}))} ≥ \frac{1 + a ∑_{i=1}^{n} sin^{2}(x_{i})}{1 + a ∑_{i=1}^{n} cos^{2}(x_{i})},

Proof. If we replace x_{i} by sin^{2}(x_{i}) for all i = 1, 2, …, n, then the corollary follow immediately from Theorem 2.2 and Corollary 2.4. □

For example, Corollary 3.1, for a = 1, yields [A, Corollary 1]. An another application for Theorem 2.3 and Corollary 2.4 we can present the following result. Corollary 3.2.

(i) For any integrable function f : [0, n] → (0, \frac{1}{2}] where n ≥ 2, and 1 ≤ a ≤ \frac{3}{2}, we have

\frac{∏_{i=1}^{n} ∫_{i−1}^{i}(1 + f(t))dt}{∏_{i=1}^{n} ∫_{i−1}^{i}(2 − f(t))dt} ≤ \frac{1 + a ∫_{0}^{n} f(t)dt}{1 + a ∫_{0}^{n}(1 − f(t))dt},

(ii) For any integrable function f : [0, n] → [\frac{1}{2}, 1) where n ≥ 2, and 1 ≤ a ≤ \frac{3}{2}, we have

\frac{∏_{i=1}^{n} ∫_{i−1}^{i}(1 + f(t))dt}{∏_{i=1}^{n} ∫_{i−1}^{i}(2 − f(t))dt} ≥ \frac{1 + a ∫_{0}^{n} f(t)dt}{1 + a ∫_{0}^{n}(1 − f(t))dt},

Proof. If setting x_{i} = ∫_{i−1}^{i} f(t)dt, then the corollary immediately follow from the Theorem 2.3 and Corollary 2.4. □

For example, Corollary 3.2, for a = 1, yields [A, Corollary 2].

References

[A] H. Alzer, Inequalities for Weierstrass products, Portugaliae Mathematica 47:1 (1990), 47–59.

[B] T.J.I’A. Bromwich, An introduction to the theory of infinite series, Macmillan, London, 1955.

[K] M.S. Klamkin, Extesions of the Weierstrass product inequalities II, Amer. Math. Monthly 82 (1975), 741–742.

[KN] M.S. Klamkin and D.J. Newman, Extesions of the Weierstrass product inequalities, Math. Mag. 43 (1970), 137– 140.

[M] D.S. Mitrinović, Analytic Inequalities, Springer Verlag, New York, 1970.

[W] C.L. Wang, Functional equation approach to inequalities II, J. Math. Anal. Appl. 78 (1980), 522–530.