[Paper Review] Refinements of Mitrinović-Cusa inequality

Zhen-Hang Yang|arXiv (Cornell University)|Jun 21, 2012

Mathematical Inequalities and Applications11 references3 citations

TL;DR

This paper refines the Mitrinović-Cusa inequality by introducing sharp bounds of the form $(\cos px)^{1/(3p^2)}$ for $\frac{\sin x}{x}$, proving that the inequality chain $(\cos x)^{1/3} < (\cos px)^{1/(3p^2)} < \frac{\sin x}{x} < (\cos qx)^{1/(3q^2)} < \frac{2+\cos x}{3}$ holds for $x \in (0, \pi/2)$ if and only if $p \in [p_1, 1)$ and $q \in (0, 1/\sqrt{5}]$, with $p_1 \approx 0.45347$. The function $p \mapsto (\cos px)^{1/(3p^2)}$ is strictly decreasing on $(0,1]$, significantly sharpening the classical inequality.

ABSTRACT

The Mitrinović-Cusa inequality states that for x\in(0,π/2) (cos x)^{1/3}

Motivation & Objective

• To refine the classical Mitrinović-Cusa inequality by introducing tighter bounds for $\frac{\sin x}{x}$ using trigonometric functions of the form $(\cos px)^{1/(3p^2)}$.
• To determine the optimal values of $p$ and $q$ such that the inequality chain $ (\cos x)^{1/3} < (\cos px)^{1/(3p^2)} < \frac{\sin x}{x} < (\cos qx)^{1/(3q^2)} < \frac{2+\cos x}{3} $ holds for all $x \in (0, \pi/2)$.
• To establish the monotonicity of the function $p \mapsto (\cos px)^{1/(3p^2)}$ on $(0,1]$, proving it is strictly decreasing.
• To provide precise estimates for integrals involving $\sin x/x$ and $\ln(\sin x)$, and to derive sharp bounds for the Catalan constant $G$ using the refined inequalities.

Proposed method

• The authors define the function $F_p(x) = \frac{\ln(\sin x / x)}{\ln(\cos px)}$ and analyze its monotonicity using series expansions of $\cot x$ and $\tan x$ involving Bernoulli numbers.
• Leveraging Lemma 1 on ratio monotonicity of differentiable functions, the paper establishes conditions under which $F_p(x)$ is increasing or decreasing, leading to bounds on $\frac{\sin x}{x}$.
• The paper uses power series expansions (Lemma 2) and integral representations of trigonometric functions to derive inequalities involving $\frac{\sin x}{x}$ and $\cos(px)$.
• By analyzing the behavior of $F_p(x)$ near $x = 0^+$ and $x = \pi/2^-$, the authors determine the optimal $p$ and $q$ for which the refined inequality chain holds.
• The method includes integration of the refined bounds over intervals such as $[0, \pi/4]$ and $[0, \pi/2]$ to estimate integrals of $\ln(\sin x)$ and $x^2 / \sin^2 x$, which are then used to bound the Catalan constant $G$.
• The paper applies the refined inequality $\frac{\sin x}{x} < (\cos qx)^{1/(3q^2)}$ with $q = 1/\sqrt{5}$ to derive tight bounds for $G$ via integral representations.

Experimental results

Research questions

• RQ1 What is the largest $p$ such that $ (\cos px)^{1/(3p^2)} < \frac{\sin x}{x} $ holds for all $x \in (0, \pi/2)$?
• RQ2 What is the smallest $q$ such that $ \frac{\sin x}{x} < (\cos qx)^{1/(3q^2)} $ holds for all $x \in (0, \pi/2)$?
• RQ3 Is the function $p \mapsto (\cos px)^{1/(3p^2)}$ strictly decreasing on $ (0,1] $?
• RQ4 Can the refined bounds be used to derive sharp estimates for integrals involving $\sin x/x$ and $\ln(\sin x)$?
• RQ5 Can the refined inequality lead to improved bounds for the Catalan constant $G$?

Key findings

• The inequality $ (\cos x)^{1/3} < (\cos px)^{1/(3p^2)} < \frac{\sin x}{x} $ holds for all $x \in (0, \pi/2)$ if and only if $p \in [p_1, 1)$, where $p_1 \approx 0.45346830977067$.
• The inequality $ \frac{\sin x}{x} < (\cos qx)^{1/(3q^2)} < \frac{2+\cos x}{3} $ holds for all $x \in (0, \pi/2)$ if and only if $q \in (0, 1/\sqrt{5}]$.
• The function $p \mapsto (\cos px)^{1/(3p^2)}$ is strictly decreasing on $ (0,1] $, which provides a continuous interpolation between the classical bounds.
• The paper establishes sharp bounds for the integral $\int_0^{\pi/4} \ln(\sin x)\,dx$ using the refined inequality, yielding $-\frac{\pi}{8}\left(2 + \ln 2 - \frac{\pi^2}{72}\right) < \int_0^{\pi/4} \ln(\sin x)\,dx < -\frac{\pi}{4}\left(2\ln 2 + 1 + \frac{\pi^2}{288} - \ln\pi\right)$.
• The paper derives three new sharp bounds for the Catalan constant $G$: $\frac{\sqrt{6}\pi}{2\sqrt{16\sqrt{3}-\pi^2}} < G < \frac{3}{32}\pi^2$, $\frac{\pi}{2}\left(\ln 2 - \ln\pi + \frac{\pi^2}{288} + 1\right) < G < \frac{\pi}{4}\left(2 - \ln 2 - \frac{\pi^2}{72}\right)$, and $\frac{\pi^2}{16} - \frac{\pi}{4}\ln 2 + \frac{8}{5}(172 - 99\sqrt{3}) < G < \frac{\pi^2}{320}(37 + 6\sqrt{3}) - \frac{\pi}{4}\ln 2$.