The Pentagonal Inequality

TL;DR

This paper derives the exact maximum value of a linear combination of five or seven cosines with positive angles summing to pi, expressed as a rational function of the coefficients, using algebraic methods.

Contribution

It provides a novel algebraic approach to determine the sharp bounds of cosine combinations with positive angles summing to pi.

Findings

Derived explicit sharp bounds as positive real fractions.

Extended the inequality analysis to five and seven cosine cases.

Used algebraic transformations to simplify the bounds expression.

Abstract

Given a positive linear combination of five (respectively seven) cosines, where the angles are positive and sum to pi, the aim of this article is to express the sharp bound of the combination as a Positive Real Fraction in the coefficients (hence cosine-free). The method uses algebraic and arithmetic manipulations with judicious transformations.

Equations

$$\sum _{i=1}^n{x}_i{x}_{i+1}cos{\alpha }_i\le cos\frac{\pi }{n}.\sum _{i=1}^n{x}_i^2$$

$$a_{1}cos{\alpha }_1+a_{2}cos{\alpha }_2+a_{3}cos{\alpha }_3+a_{4}cos{\alpha }_4+a_{5}cos{\alpha }_5$$

$$\le$$

$$\big{(}\frac{1+\sqrt{5}}{4})\hskip 1.42271pt\frac{1}{a_{1}a_{2}a_{3}a_{4}a_{5}}\hskip 1.42271pt(a_{1}^{2}a_{5}^{2}a_{2}^{2}+a_{5}^{2}a_{2}^{2}a_{3}^{2}+a_{2}^{2}a_{3}^{2}a_{4}^{2}+a_{3}^{2}a_{4}^{2}a_{1}^{2}+a_{4}^{2}a_{1}^{2}a_{5}^{2})$$

$$a_{1}cos{\alpha }_1+a_{2}cos{\alpha }_2+a_{3}cos{\alpha }_3+a_{4}cos{\alpha }_4+a_{5}cos{\alpha }_5$$

$$\le$$

$$(\frac{1+\sqrt{5}}{4})\frac{1}{a_1{a}_2{a}_3{a}_4{a}_5}(a_1^2{a}_2^2{a}_3^2+a_2^2{a}_3^2{a}_4^2+a_3^2{a}_4^2{a}_5^2+a_4^2{a}_5^2{a}_1^2+a_5^2{a}_1^2{a}_2^2)$$

$$a_{1}sin{\alpha }_1=a_{2}sin{\alpha }_2=a_{3}sin{\alpha }_3=a_{4}sin{\alpha }_4=a_{5}sin{\alpha }_5.$$

$$\varphi (x_1,x_2,x_3,x_4,x_5)=x_1{x}_2{x}_3+x_2{x}_3{x}_4+x_3{x}_4{x}_5+x_4{x}_5{x}_1+x_5{x}_1{x}_2.$$

$$\varphi (a_1,a_2,a_3,a_4,a_5)-\varphi ({\sigma }_0)$$

$$=a_3(a_4-a_2)(a_5-a_1)\ge 0$$

$$\varphi (a_1,a_2,a_3,a_5,a_4)-\varphi ({\sigma }_0)$$

$$=a_1(a_3-a_2)(a_5-a_4)+a_3(a_4-a_1)(a_5-a_2)\ge 0$$

$$\varphi (a_1,a_2,a_4,a_3,a_5)-\varphi ({\sigma }_0)$$

$$=a_1(a_3-a_2)(a_5-a_4)+a_5(a_3-a_1)(a_4-a_2)\ge 0$$

$$\varphi (a_1,a_2,a_4,a_5,a_3)-\varphi ({\sigma }_0)=\varphi (a_1,a_3,a_5,a_4,a_2)-\varphi ({\sigma }_0)$$

$$=a_1(a_3-a_2)(a_5-a_4)+a_3(a_4-a_1)(a_5-a_2)+a_5(a_2-a_1)(a_4-a_3)\ge 0$$

$$\varphi (a_1,a_2,a_5,a_3,a_4)-\varphi ({\sigma }_0)$$

$$=a_4(a_3-a_1)(a_5-a_2)\ge 0$$

$$\varphi (a_1,a_2,a_5,a_4,a_3)-\varphi ({\sigma }_0)$$

$$=a_3(a_4-a_1)(a_5-a_2)+a_5(a_2-a_1)(a_4-a_3)\ge 0$$

$$\varphi (a_1,a_3,a_2,a_4,a_5)-\varphi ({\sigma }_0)$$

$$=a_1(a_3-a_2)(a_5-a_4)+a_2(a_4-a_3)(a_5-a_1)\ge 0$$

$$\varphi (a_1,a_3,a_2,a_5,a_4)-\varphi ({\sigma }_0)$$

$$=a_2(a_4-a_1)(a_5-a_3)\ge 0$$

$$\varphi (a_1,a_3,a_4,a_2,a_5)-\varphi ({\sigma }_0)$$

$$=a_5(a_2-a_1)(a_4-a_3)\ge 0$$

$$\varphi (a_1,a_3,a_5,a_2,a_4)-\varphi ({\sigma }_0)$$

$$=a_1(a_3-a_2)(a_5-a_4)+a_4(a_2-a_1)(a_5-a_3)\ge 0$$

$$\varphi (a_1,a_4,a_2,a_3,a_5)-\varphi ({\sigma }_0)=\varphi (a_1,a_5,a_3,a_2,a_4)-\varphi ({\sigma }_0)=a_1(a_3-a_2)(a_5-a_4)\ge 0$$

$$\varphi (a_1,a_4,a_3,a_2,a_5)-\varphi ({\sigma }_0)=\varphi (a_1,a_5,a_2,a_3,a_4)-\varphi ({\sigma }_0)=0$$

$$\sum\limits_{i=1}^{5}b_{i}cos\alpha_{i}\hskip 2.84544pt\leq\hskip 2.84544pt\displaystyle{\big{(}\frac{1+\sqrt{5}}{4})\hskip 1.42271pt\frac{1}{b_{1}b_{2}b_{3}b_{4}b_{5}}\hskip 1.42271pt\phi(b_{1}^{2}b_{2}^{2}b_{3}^{2}b_{4}^{2}b_{5}^{2})}$$

$$\sum _{i=1}^5{x}_i^2=P^{-1}\varphi (b_1^2,b_2^2,b_3^2,b_4^2,b_5^2)$$

$$\sum _{i=1}^5{x}_i{x}_{i+1}cos{\beta }_i=\sum _{i=1}^5{b}_{i}cos{\alpha }_i$$

$$\sum _{i=1}^5{x}_i{x}_{i+1}cos{\alpha }_i\le cos\frac{\pi }{5}.\sum _{i=1}^5{x}_i^2$$

Full text

The Pentagonal Inequality

Roy Barbara

1. Introduction

We recall the following result, conjectured by L. Fejes Tóth (see [References]) and proved by Lenhard H-C (see [References]).

Theorem 0**.**

(Tóth’s inequality) Let $n\geq 3$. Let $x_{1},x_{2},...,x_{n}$ and $\alpha_{1},\alpha_{2},...,\alpha_{n}$ be positive real numbers with $\sum\limits_{i=1}^{n}\alpha_{i}=\pi.$ Then, with $x_{n+1}=x_{1}$, we have the following inequality:

[TABLE]

Our interest lies in a sharp majoration of a sum as $\sum\limits_{i=1}^{n}a_{i}cos\alpha_{i},$ where $a_{i},\alpha_{i}>0$ and $\sum\limits_{i=1}^{n}\alpha_{i}=\pi.$ Results for the cases $n=3,4$ can be found in [References]. In this article, we focus our attention on the case $n=5$. In section 4, we highlight the sharpness of the result. In section 5, we quickly study the case $n=7$. The case $n=6$ remains open.

2. The Results

We present the Pentagonal Inequality in two forms: the strong form (theorem 1) and the normal form (theorem 2). Though theorem 1 is more precise, theorem 2 remains sometimes more practical.

Theorem 1**.**

Let $a_{1},a_{2},...,a_{5}$ and $\alpha_{1},\alpha_{2},...,\alpha_{5}$ be positive real numbers with $\sum\limits_{i=1}^{5}\alpha_{i}=\pi$ and such that $a_{1}\leq a_{2}\leq a_{3}\leq a_{4}\leq a_{5}.$   Then, we have the sharp inequality:

[TABLE]

[TABLE]

[TABLE]

Theorem 2**.**

Let $a_{1},a_{2},...,a_{5}$ and $\alpha_{1},\alpha_{2},...,\alpha_{5}$ be positive real numbers with $\sum\limits_{i=1}^{5}\alpha_{i}=\pi$. Then, we have:

[TABLE]

[TABLE]

[TABLE]

We briefly indicate that the case of equality in either theorem 1 or 2 requires the condition:

[TABLE]

3. Proofs of Theorems 1 and 2

Let $a_{1},a_{2},...,a_{5}$ be positive real numbers. A circular permutation of the $a_{i}$ can be represented by $\sigma=(x_{1},x_{2},...,x_{5})$, with $x_{1}=a_{1}$, and where $x_{2},x_{3},x_{4},x_{5}$ are $a_{2},a_{3},a_{4},a_{5}$ in some order. There are $4!=24$ such permutations.

In all what follows, we use the following notation:

Notation: Let $\sigma=(x_{1},x_{2},...,x_{5})$ be a circular permutation of $a_{1},a_{2},...,a_{5}$ (with $x_{1}=a_{1}$). We define $\phi(\sigma)$, also denoted by $\phi(x_{1},x_{2},...,x_{5})$ as follows:

[TABLE]

Lemma 1**.**

Let $a_{1}\leq a_{2}\leq...\leq a_{5}$ be positive real numbers. Then, there is a circular permutation of the $a_{i}$, say $\sigma_{0}$,  such that $\phi(\sigma_{0})$ minimizes all the $\phi(\sigma)$. Namely, $\sigma_{0}$ is the cycle  $\sigma_{0}=(a_{1},a_{5},a_{2},a_{3},a_{4}).$

Equivalently, one might take $\sigma_{1}=(a_{1},a_{4},a_{3},a_{2},a_{5})$   since $\phi(\sigma_{1})=\phi(\sigma_{0}).$

Proof.

After some amount of algebra, we find the following identities (and the result follows):

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

∎

Lemma 2**.**

Let $a_{1},a_{2},...,a_{5}$ and $\alpha_{1},\alpha_{2},...,\alpha_{5}$ be positive real numbers with $\sum\limits_{i=1}^{5}\alpha_{i}=\pi$. If $\sigma=(b_{1},b_{2},...,b_{5})$ is any circular permutation of the $a_{i}$ (with $b_{1}=a_{1}$), then we have:

[TABLE]

Proof.

Set $\beta_{1}=\alpha_{1},\hskip 2.84544pt\beta_{2}=\alpha_{4},\hskip 2.84544pt\beta_{3}=\alpha_{2},\hskip 2.84544pt\beta_{4}=\alpha_{5},$ and $\beta_{5}=\alpha_{3}$ (so $\sum\limits_{i=1}^{5}\beta_{i}=\pi).$

Set $P=b_{1}b_{2}b_{3}b_{4}b_{5}=a_{1}a_{2}a_{3}a_{4}a_{5},$  and set

$\displaystyle{x_{1}=\sqrt{\frac{b_{1}b_{2}b_{3}}{b_{4}b_{5}}},\hskip 8.5359ptx_{2}=\sqrt{\frac{b_{1}b_{4}b_{5}}{b_{2}b_{3}}},\hskip 8.5359ptx_{3}=\sqrt{\frac{b_{2}b_{3}b_{4}}{b_{1}b_{5}}},\hskip 8.5359ptx_{4}=\sqrt{\frac{b_{1}b_{2}b_{5}}{b_{3}b_{4}}}},\hskip 8.5359ptx_{5}=\sqrt{\frac{b_{3}b_{4}b_{5}}{b_{1}b_{2}}}.$

We have $x_{1}^{2}=P^{-1}(b_{1}^{2}b_{2}^{2}b_{3}^{2}),\hskip 2.84544ptx_{3}^{2}=P^{-1}(b_{2}^{2}b_{3}^{2}b_{4}^{2}),\hskip 2.84544ptx_{5}^{2}=P^{-1}(b_{3}^{2}b_{4}^{2}b_{5}^{2}),$  $x_{2}^{2}=P^{-1}(b_{4}^{2}b_{5}^{2}b_{1}^{2}),$ and $x_{4}^{2}=P^{-1}(b_{5}^{2}b_{1}^{2}b_{2}^{2})$. Summing, we obtain:

[TABLE]

On the other hand we have: $x_{1}x_{2}cos\beta_{1}=b_{1}cos\alpha_{1},\hskip 2.84544ptx_{2}x_{3}cos\beta_{2}=b_{4}cos\alpha_{4},\hskip 2.84544pt$ $x_{3}x_{4}cos\beta_{3}=b_{2}cos\alpha_{2},\hskip 2.84544pt$ $x_{4}x_{5}cos\beta_{4}=b_{5}cos\alpha_{5},$  and  $x_{5}x_{1}cos\beta_{5}=b_{3}cos\alpha_{3}.$

Hence, with $x_{6}=x_{1}$, we get:

[TABLE]

By theorem ‣ 1, we have $\sum\limits_{i=1}^{5}x_{i}x_{i+1}cos\beta_{i}\leq cos\frac{\pi}{5}.\sum\limits_{i=1}^{5}x_{i}^{2}$. From this, (2), (3), and $\displaystyle{cos\frac{\pi}{5}=\frac{1+\sqrt{5}}{4}}$, we obtain (1).

∎

Proofs of theorems 1 and 2: To get theorem 2, just apply lemma 2 to $\sigma=(a_{1},a_{2},a_{3},a_{4},a_{5}).$ To get theorem 1, apply lemma 2 to the cycle $\sigma_{0}=(a_{1},a_{5},a_{2},a_{3},a_{4}),$ obtaining, in virtue of lemma 1, the sharpest form of the Pentagonal Inequality.

Note the following: If $a_{1}<a_{2}<...<a_{5}$ (or more simply if $a_{1}\leq a_{2}\leq...\leq a_{5}$ and $a_{2}<a_{4})$, then, $\phi(a_{1}^{2},a_{5}^{2},a_{2}^{2},a_{3}^{2},a_{4}^{2})<\phi(a_{1}^{2},a_{2}^{2},a_{3}^{2},a_{4}^{2},a_{5}^{2}).$ Hence, in this case, while equality in theorem 1 might be reached, equality in theorem 2 never arises.

4. Sharpness of the Pentagonal Inequality

To see why the Pentagonal Inequality is sharp, we prove that Tóth’s inequality for $n=5$ can be derived from the Pentagonal Inequality (the normal form suffices): Indeed, let $x_{1},x_{2},...,x_{5}$ and $\alpha_{1},\alpha_{2},...,\alpha_{5}$ be positive real numbers with $\sum\limits_{i=1}^{5}\alpha_{i}=\pi$. Given by hypothesis that theorem 2 holds, we prove (with $x_{6}=x_{1}$) that

[TABLE]

Set $\beta_{1}=\alpha_{1},\hskip 2.84544pt\beta_{2}=\alpha_{3},\hskip 2.84544pt\beta_{3}=\alpha_{5},\hskip 2.84544pt\beta_{4}=\alpha_{2},$  and  $\beta_{5}=\alpha_{4}$ (so $\sum\limits_{i=1}^{5}\beta_{i}=\pi$). Set also $a_{1}=x_{1}x_{2},\hskip 2.84544pta_{2}=x_{3}x_{4},\hskip 2.84544pt$ $a_{3}=x_{5}x_{1},\hskip 2.84544pta_{4}=x_{2}x_{3},$  and  $a_{5}=x_{4}x_{5}.$

By theorem 2, we have:

[TABLE]

We have: $a_{1}cos\beta_{1}=x_{1}x_{2}cos\alpha_{1},\hskip 2.84544pta_{2}cos\beta_{2}=x_{3}x_{4}cos\alpha_{3},$ $a_{3}cos\beta_{3}=x_{5}x_{1}cos\alpha_{5},\hskip 2.84544pta_{4}cos\beta_{4}=x_{2}x_{3}cos\alpha_{2},$  and $a_{5}cos\beta_{5}=x_{4}x_{5}cos\alpha_{4}.$  Hence,

[TABLE]

Set  $P=a_{1}a_{2}a_{3}a_{4}a_{5}.$   We easily find that   $P^{-1}\vspace{0.07cm}(a_{1}^{2}a_{2}^{2}a_{3}^{2})=x_{1}^{2},\hskip 4.26773ptP^{-1}(a_{2}^{2}a_{3}^{2}a_{4}^{2})=x_{3}^{2},\hskip 4.26773ptP^{-1}(a_{3}^{2}a_{4}^{2}a_{5}^{2})=x_{5}^{2},$

$P^{-1}(a_{4}^{2}a_{5}^{2}a_{1}^{2})=x_{2}^{2},$ and $P^{-1}(a_{5}^{2}a_{1}^{2}a_{2}^{2})=x_{4}^{2}.$

Summing yields $\sum\limits_{i=1}^{5}x_{i}^{2}=P^{-1}\phi(a_{1}^{2},a_{2}^{2},a_{3}^{2},a_{4}^{2},a_{5}^{2}).$ Hence the right-side in (5) is $cos\frac{\pi}{5}.\sum\limits_{i=1}^{5}x_{i}^{2}.$ From this, (5), and (6), we get (4).

5. The Heptagonal Inequality

The case of an odd integer $n\geq 7$ can be treated similarly. For $n=7$, we obtain:

Theorem 3**.**

(the normal form) Let $a_{1},a_{2},...,a_{7}$ and $\alpha_{1},\alpha_{2},...,\alpha_{7}$ be positive real numbers with $\sum\limits_{i=1}^{7}\alpha_{i}=\pi.$ Then, we have:

[TABLE]

where

[TABLE]

Proof.

Set $\beta_{1}=\alpha_{1},\hskip 2.84544pt\beta_{2}=\alpha_{5},\hskip 2.84544pt\beta_{3}=\alpha_{2},\hskip 2.84544pt\beta_{4}=\alpha_{6},\hskip 2.84544pt\beta_{5}=\alpha_{3},\hskip 2.84544pt\beta_{6}=\alpha_{7},\hskip 2.84544pt\beta_{7}=\alpha_{4},$  and set

$\displaystyle{x_{1}=\sqrt{\frac{a_{1}a_{2}a_{3}a_{4}}{a_{5}a_{6}a_{7}}},\hskip 4.26773ptx_{2}=\sqrt{\frac{a_{1}a_{5}a_{6}a_{7}}{a_{2}a_{3}a_{4}}},\hskip 4.26773ptx_{3}=\sqrt{\frac{a_{2}a_{3}a_{4}a_{5}}{a_{1}a_{6}a_{7}}},\hskip 4.26773ptx_{4}=\sqrt{\frac{a_{1}a_{2}a_{6}a_{7}}{a_{3}a_{4}a_{5}}}},$ $\displaystyle{\hskip 4.26773ptx_{5}=\sqrt{\frac{a_{3}a_{4}a_{5}a_{6}}{a_{1}a_{2}a_{7}}}},$

$\vspace{0.2cm}\displaystyle{x_{6}=\sqrt{\frac{a_{1}a_{2}a_{3}a_{7}}{a_{4}a_{5}a_{6}}},\hskip 2.84544pt\text{and }\hskip 2.84544ptx_{7}=\sqrt{\frac{a_{4}a_{5}a_{6}a_{7}}{a_{1}a_{2}a_{3}}}}.$

Then, apply theorem ‣ 1 with these values of $x_{1},x_{2},...,x_{7}$ and $\beta_{1},\beta_{2},...,\beta_{7}.$ ∎