Arithmetic, Geometric, and Harmonic Means Inequality
Definitions
For $n$ positive numbers ${x}_1,{x}_2,\cdots,{x}_n$, the arithmetic mean, geometric mean, and harmonic mean are defined as:
- Arithmetic Mean : $$ \sum_{ k=1 }^{ n }{ \frac { {x}_k }{ n } }=\frac { {x}_1+{x}_2+\cdots+{x}_n }{ n } $$
- Geometric Mean : $$ \prod_{ k=1 }^{ n }{ { {x}_k }^{ \frac { 1 }{ n } } }=\sqrt [ n ]{ {x}_1{x}_2\cdots{x}_n } $$
- Harmonic Mean : $$ \left( \frac { \sum_{ k=1 }^{ n }{ \frac { 1 }{ {x}_k } } }{ n } \right)^{-1}=\frac { n }{ \frac { 1 }{ {x}_1 }+\frac { 1 }{ {x}_2 }+\cdots+\frac { 1 }{ {x}_n } } $$
Theorem
The following inequality holds for these means:
$$ \frac { {x}_1+{x}_2+\cdots+{x}_n }{ n }\ge \sqrt [ n ]{ {x}_1{x}_2\cdots{x}_n }\ge \frac { n }{ \frac { 1 }{ {x}_1 }+\frac { 1 }{ {x}_2 }+\cdots+\frac { 1 }{ {x}_n } } $$
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 $n=2$, 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 $f$ is a convex function and $E(X) < \infty$, then the following inequality holds: $$ E{f(X)}\ge f{E(X)} $$
Arithmetic-Geometric
Let $f(x)=-\ln x$, then $f$ is convex on the interval $(0,\infty )$. Assume that a random variable $X$ has the probability mass function
$$ p(X=x)=\begin{cases}{1 \over n} & , x={x}_1,{x}_2, \cdots ,{x}_n \\ 0 & , \text{otherwise}\end{cases} $$
Then $E(X)$ is
$$ \frac { {x}_1+{x}_2+…+{x}_n }{ n }<\infty $$
hence finite. This satisfies all necessary conditions for Jensen’s inequality, yielding:
$$ E(-\ln X)\ge –\ln E(X) $$
The left-hand side is
$$ \begin{align*} E(-\ln X)&=-E(\ln X) \\ &=-\frac { 1 }{ n } \sum_{ k=1 }^{ n }{ \ln{x}_k } \\ &=-\frac { 1 }{ n }\ln \prod_{ k=1 }^{ n }{ {x}_k } \\ &=-\ln { \left( \prod_{ k=1 }^{ n }{ {x}_k } \right) }^{ \frac { 1 }{ n } } \\ &=-\ln\prod_{ k=1 }^{ n }{ { {x}_k }^{ \frac { 1 }{ n } } } \end{align*} $$
The right-hand side is
$$ \begin{align*} -\ln E(X)=-\ln\frac { 1 }{ n }\sum_{ k=1 }^{ n }{ {x}_k } \end{align*} $$
Upon rearranging, we get
$$ \begin{align*} -\ln\prod_{ k=1 }^{ n }{ { {x}_k }^{ \frac { 1 }{ n } } } \ge& -\ln\frac { 1 }{ n }\sum_{ k=1 }^{ n }{ {x}_k } \\ \implies \ln\frac { 1 }{ n }\sum_{ k=1 }^{ n }{ {x}_k } \ge& \ln\prod_{ k=1 }^{ n }{ { {x}_k }^{ \frac { 1 }{ n } } } \\ \implies \frac { 1 }{ n }\sum_{ k=1 }^{ n }{ {x}_k } \ge& \prod_{ k=1 }^{ n }{ { {x}_k }^{ \frac { 1 }{ n } } } \\ \implies \frac { {x}_1+{x}_2+…+{x}_n }{ n } \ge& \sqrt [ n ]{ {x}_1{x}_2…{x}_n } \end{align*} $$
■
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
$$ \frac { {x}_1+{x}_2+…+{x}_n }{ n }\ge \sqrt [ n ]{ {x}_1{x}_2…{x}_n } $$
By setting $\displaystyle {x}_k=\frac { 1 }{ {y}_k }$, we get
$$ \begin{align*} \frac { \frac { 1 }{ {y}_1 }+\frac { 1 }{ {y}_2 }+…+\frac { 1 }{ {y}_n } }{ n }\ge \sqrt [ n ]{ \frac { 1 }{ {y}_1 }\frac { 1 }{ {y}_2 }…\frac { 1 }{ {y}_n } } \\ \implies \frac { 1 }{ \sqrt [ n ]{ \frac { 1 }{ {y}_1 }\frac { 1 }{ {y}_2 }…\frac { 1 }{ {y}_n } } }\ge \frac { n }{ \frac { 1 }{ {y}_1 }+\frac { 1 }{ {y}_2 }+…+\frac { 1 }{ n{y}_n } } \\ \implies \sqrt [ n ]{ {y}_1{y}_2…{y}_n }\ge \frac { n }{ \frac { 1 }{ n{y}_1 }+\frac { 1 }{ n{y}_2 }+…+\frac { 1 }{ n{y}_n } } \end{align*} $$
■