[Paper Review] A new proof of the geometric-arithmetic mean inequality by Cauchy's integral formula
This paper presents a novel proof of the geometric-arithmetic mean inequality using Cauchy's integral formula in complex analysis. By deriving an integral representation for the principal branch of the geometric mean $ G_n(a+z) $, the authors show that the geometric mean is bounded above by the arithmetic mean, with equality iff all $ a_k $ are equal, leveraging the non-negativity of the integral remainder term.
Let $a=(a_1,a_2,...c,a_n)$ for $n\in\mathbb{N}$ be a given sequence of positive numbers. In the paper, the authors establish, by using Cauchy's integral formula in the theory of complex functions, an integral representation of the principal branch of the geometric mean {equation*} G_n(a+z)=\Biggl[\prod_{k=1}^n(a_k+z)\Biggr]^{1/n} {equation*} for $z\in\mathbb{C}\setminus(-\infty,-\min\{a_k,1\le k\le n\}]$, and then provide a new proof of the well known GA mean inequality.
Motivation & Objective
- To establish a new proof of the classical geometric-arithmetic mean inequality using complex analysis.
- To derive an integral representation for the principal branch of the geometric mean $ G_n(a+z) $ via Cauchy's integral formula.
- To demonstrate the GA inequality as a direct consequence of the integral representation, with equality iff all $ a_k $ are equal.
- To provide a complex-analytic framework for understanding the GA mean inequality through boundary behavior and imaginary part limits.
Proposed method
- Derive an integral representation of $ G_n(a+z) $ using Cauchy’s integral formula in the complex plane, valid for $ z \in \mathbb{C} \setminus (-\infty, -a_1] $.
- Define the auxiliary function $ f_n(z) = G_n(a+z) - z $, whose limit at infinity is $ A_n(a) $, establishing a link to the arithmetic mean.
- Use the boundary behavior of the imaginary part of $ h_n(z) = G_n(a - a_1 + z) - z $ on the negative real axis to extract jump discontinuities related to the geometric mean.
- Apply contour integration over a keyhole-shaped path avoiding the branch cut on $ (-\infty, -a_1] $, exploiting analyticity and decay at infinity.
- Take limits as $ \varepsilon \to 0^+ $ and $ r \to \infty $ to evaluate the contour integrals, yielding a real integral expression for $ f_n(z) $.
- Substitute $ z = 0 $ into the final integral representation to obtain $ G_n(a) \leq A_n(a) $, proving the GA inequality.
Experimental results
Research questions
- RQ1Can the geometric-arithmetic mean inequality be proven using complex analysis and Cauchy’s integral formula?
- RQ2What is the integral representation of the principal branch of the geometric mean $ G_n(a+z) $ in the complex plane?
- RQ3How does the imaginary part of the complex function $ h_n(z) $ behave on the negative real axis, and what does it reveal about the geometric mean?
- RQ4What role does the limit $ \lim_{z \to \infty} f_n(z) = A_n(a) $ play in the proof structure?
- RQ5Under what conditions does equality hold in the GA inequality, and how is this reflected in the integral representation?
Key findings
- The geometric mean $ G_n(a) $ admits the integral representation $ G_n(a) = A_n(a) - \frac{1}{\pi} \sum_{\ell=1}^{n-1} \sin\frac{\ell\pi}{n} \int_{a_\ell}^{a_{\ell+1}} \left| \prod_{k=1}^n (a_k - t) \right|^{1/n} \frac{dt}{t} $.
- The integral remainder term is non-negative for all $ a_k > 0 $, implying $ G_n(a) \leq A_n(a) $, with equality iff all $ a_k $ are equal.
- The imaginary part of $ h_n(-t + i\varepsilon) $ exhibits a jump discontinuity proportional to $ \sin(\ell\pi/n) $ on intervals $ (a_\ell - a_1, a_{\ell+1} - a_1] $, which is crucial for the integral derivation.
- The limit $ \lim_{z \to \infty} f_n(z) = A_n(a) $ is rigorously justified using L’Hôpital’s rule in the complex setting, linking the function to the arithmetic mean.
- The proof establishes a new connection between classical inequalities and complex analysis, specifically through contour integration and boundary value analysis.
- The equality case $ a_1 = a_2 = \cdots = a_n $ is characterized by the vanishing of the integral remainder, as the integrand becomes zero.
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This review was created by AI and reviewed by human editors.