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Arithmetic, Geometric, and Harmonic Means Inequality 📂Lemmas

Arithmetic, Geometric, and Harmonic Means Inequality

Definitions

For nn positive numbers x1,x2,⋯ ,xn{x}_1,{x}_2,\cdots,{x}_n, the arithmetic mean, geometric mean, and harmonic mean are defined as:

  • Arithmetic Mean : ∑k=1nxkn=x1+x2+⋯+xnn \sum_{ k=1 }^{ n }{ \frac { {x}_k }{ n } }=\frac { {x}_1+{x}_2+\cdots+{x}_n }{ n }
  • Geometric Mean : ∏k=1nxk1n=x1x2⋯xnn \prod_{ k=1 }^{ n }{ { {x}_k }^{ \frac { 1 }{ n } } }=\sqrt [ n ]{ {x}_1{x}_2\cdots{x}_n }
  • Harmonic Mean : (∑k=1n1xkn)−1=n1x1+1x2+⋯+1xn \left( \frac { \sum_{ k=1 }^{ n }{ \frac { 1 }{ {x}_k } } }{ n } \right)^{-1}=\frac { n }{ \frac { 1 }{ {x}_1 }+\frac { 1 }{ {x}_2 }+\cdots+\frac { 1 }{ {x}_n } }

Theorem

The following inequality holds for these means:

x1+x2+⋯+xnn≥x1x2⋯xnn≥n1x1+1x2+⋯+1xn \frac { {x}_1+{x}_2+\cdots+{x}_n }{ n }\ge \sqrt [ n ]{ {x}_1{x}_2\cdots{x}_n }\ge \frac { n }{ \frac { 1 }{ {x}_1 }+\frac { 1 }{ {x}_2 }+\cdots+\frac { 1 }{ {x}_n } }

Explanation

High school students might have heard about the arithmetic-geometric mean at some point. It is not typically defined by a specific name but is commonly passed down colloquially as “Arith-Geo.” For the case when n=2n=2, its proof is simple and useful even for high school level problem solving. A general proof at the high school level requires the intervention of messy expressions using mathematical induction, but instead, a more sophisticated but challenging proof is introduced.

Proof

Strategy: Utilizing the following lemma:

Jensen’s Inequality: If ff is a convex function and E(X)<∞E(X) < \infty, then the following inequality holds: Ef(X)≥fE(X) E{f(X)}\ge f{E(X)}

Arithmetic-Geometric

Let f(x)=−ln⁡xf(x)=-\ln x, then ff is convex on the interval (0,∞)(0,\infty ). Assume that a random variable XX has the probability mass function

p(X=x)={1n,x=x1,x2,⋯ ,xn0,otherwise p(X=x)=\begin{cases}{1 \over n} & , x={x}_1,{x}_2, \cdots ,{x}_n \\ 0 & , \text{otherwise}\end{cases}

Then E(X)E(X) is

x1+x2+…+xnn<∞ \frac { {x}_1+{x}_2+…+{x}_n }{ n }<\infty

hence finite. This satisfies all necessary conditions for Jensen’s inequality, yielding:

E(−ln⁡X)≥–ln⁡E(X) E(-\ln X)\ge –\ln E(X)

The left-hand side is

E(−ln⁡X)=−E(ln⁡X)=−1n∑k=1nln⁡xk=−1nln⁡∏k=1nxk=−ln⁡(∏k=1nxk)1n=−ln⁡∏k=1nxk1n \begin{align*} E(-\ln X)&=-E(\ln X) \\ &=-\frac { 1 }{ n } \sum_{ k=1 }^{ n }{ \ln{x}_k } \\ &=-\frac { 1 }{ n }\ln \prod_{ k=1 }^{ n }{ {x}_k } \\ &=-\ln { \left( \prod_{ k=1 }^{ n }{ {x}_k } \right) }^{ \frac { 1 }{ n } } \\ &=-\ln\prod_{ k=1 }^{ n }{ { {x}_k }^{ \frac { 1 }{ n } } } \end{align*}

The right-hand side is

−ln⁡E(X)=−ln⁡1n∑k=1nxk \begin{align*} -\ln E(X)=-\ln\frac { 1 }{ n }\sum_{ k=1 }^{ n }{ {x}_k } \end{align*}

Upon rearranging, we get

−ln⁡∏k=1nxk1n≥−ln⁡1n∑k=1nxk  ⟹  ln⁡1n∑k=1nxk≥ln⁡∏k=1nxk1n  ⟹  1n∑k=1nxk≥∏k=1nxk1n  ⟹  x1+x2+…+xnn≥x1x2…xnn \begin{align*} -\ln\prod_{ k=1 }^{ n }{ { {x}_k }^{ \frac { 1 }{ n } } } \ge& -\ln\frac { 1 }{ n }\sum_{ k=1 }^{ n }{ {x}_k } \\ \implies \ln\frac { 1 }{ n }\sum_{ k=1 }^{ n }{ {x}_k } \ge& \ln\prod_{ k=1 }^{ n }{ { {x}_k }^{ \frac { 1 }{ n } } } \\ \implies \frac { 1 }{ n }\sum_{ k=1 }^{ n }{ {x}_k } \ge& \prod_{ k=1 }^{ n }{ { {x}_k }^{ \frac { 1 }{ n } } } \\ \implies \frac { {x}_1+{x}_2+…+{x}_n }{ n } \ge& \sqrt [ n ]{ {x}_1{x}_2…{x}_n } \end{align*}

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This proves the inequality between the arithmetic and geometric means. Using this, let’s prove the inequality between the geometric and harmonic means.

Geometric-Harmonic

x1+x2+…+xnn≥x1x2…xnn \frac { {x}_1+{x}_2+…+{x}_n }{ n }\ge \sqrt [ n ]{ {x}_1{x}_2…{x}_n }

By setting xk=1yk\displaystyle {x}_k=\frac { 1 }{ {y}_k }, we get

1y1+1y2+…+1ynn≥1y11y2…1ynn  ⟹  11y11y2…1ynn≥n1y1+1y2+…+1nyn  ⟹  y1y2…ynn≥n1ny1+1ny2+…+1nyn \begin{align*} \frac { \frac { 1 }{ {y}_1 }+\frac { 1 }{ {y}_2 }+…+\frac { 1 }{ {y}_n } }{ n }\ge \sqrt [ n ]{ \frac { 1 }{ {y}_1 }\frac { 1 }{ {y}_2 }…\frac { 1 }{ {y}_n } } \\ \implies \frac { 1 }{ \sqrt [ n ]{ \frac { 1 }{ {y}_1 }\frac { 1 }{ {y}_2 }…\frac { 1 }{ {y}_n } } }\ge \frac { n }{ \frac { 1 }{ {y}_1 }+\frac { 1 }{ {y}_2 }+…+\frac { 1 }{ n{y}_n } } \\ \implies \sqrt [ n ]{ {y}_1{y}_2…{y}_n }\ge \frac { n }{ \frac { 1 }{ n{y}_1 }+\frac { 1 }{ n{y}_2 }+…+\frac { 1 }{ n{y}_n } } \end{align*}

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