Inequality Between Arithmetic, Geometric, and Harmonic Means
Definition
For $n$ positive numbers ${x}_{1},{x}_{2},\cdots,{x}_{n}$, the arithmetic, geometric, and harmonic means are as follows.
- 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
Regarding these, the following inequality holds.
$$ \frac { {x}_{1}+{x}_{2}+\cdots+{x}_{n} }{ n }\ge \sqrt [ n ]{ {x}_{1}{x}_{2}\cdots{x}_{n} }\ge \frac { n }{ \frac { 1 }{ {x}_{1} }+\frac { 1 }{ {x}_{2} }+\cdots+\frac { 1 }{ {x}_{n} } } $$
Explanation
If you are a high school student, you will hear the term arithmetic-geometric mean at least once, but it is rarely pinned down as a formal name; it is usually passed around by the abbreviation “arithmetic-geometric.” For the case $n=2$, the proof is simple and it is also useful for problem solving at the high school level. At the high school level, the general proof inevitably relies on mathematical induction, which requires handling messy expressions; here we introduce a proof that is harder than that, but more elegant.
Proof
Strategy: We use 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 a convex function on the interval $(0,\infty )$. Suppose the random variable $X$ has the probability mass function
$$ p(X=x)=\begin{cases}{1 \over n} & , x={x}_{1},{x}_{2}, \cdots ,{x}_{n} \\ 0 & , \text{그 이외의 경우}\end{cases} $$
Then $E(X)$ is
$$ \frac { {x}_{1}+{x}_{2}+…+{x}_{n} }{ n }<\infty $$
and is therefore finite. Thus all the conditions required by Jensen’s inequality are satisfied, and we obtain the following.
$$ E(-\ln X)\ge –\ln E(X) $$
Here, the left-hand side is
$$ \begin{align*} E(-\ln X)&=-E(\ln X) \\ &=-\frac { 1 }{ n } \sum _{ k=1 }^{ n }{ \ln{x}_{k} } \\ &=-\frac { 1 }{ n }\ln \prod _{ k=1 }^{ n }{ {x}_{k} } \\ &=-\ln { \left( \prod _{ k=1 }^{ n }{ {x}_{k} } \right) }^{ \frac { 1 }{ n } } \\ &=-\ln\prod _{ k=1 }^{ n }{ { {x}_{k} }^{ \frac { 1 }{ n } } } \end{align*} $$
and the right-hand side is
$$ \begin{align*} -\ln E(X)=-\ln\frac { 1 }{ n }\sum _{ k=1 }^{ n }{ {x}_{k} } \end{align*} $$
Rearranging this again gives
$$ \begin{align*} -\ln\prod _{ k=1 }^{ n }{ { {x}_{k} }^{ \frac { 1 }{ n } } } \ge& -\ln\frac { 1 }{ n }\sum _{ k=1 }^{ n }{ {x}_{k} } \\ \implies \ln\frac { 1 }{ n }\sum _{ k=1 }^{ n }{ {x}_{k} } \ge& \ln\prod _{ k=1 }^{ n }{ { {x}_{k} }^{ \frac { 1 }{ n } } } \\ \implies \frac { 1 }{ n }\sum _{ k=1 }^{ n }{ {x}_{k} } \ge& \prod _{ k=1 }^{ n }{ { {x}_{k} }^{ \frac { 1 }{ n } } } \\ \implies \frac { {x}_{1}+{x}_{2}+…+{x}_{n} }{ n } \ge& \sqrt [ n ]{ {x}_{1}{x}_{2}…{x}_{n} } \end{align*} $$
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This proves that the inequality between the arithmetic mean and the geometric mean holds. Using this, let us prove that the inequality between the geometric mean and the harmonic mean holds.
Geometric-Harmonic
In
$$ \frac { {x}_{1}+{x}_{2}+…+{x}_{n} }{ n }\ge \sqrt [ n ]{ {x}_{1}{x}_{2}…{x}_{n} } $$
set $\displaystyle {x}_{k}=\frac { 1 }{ {y}_{k} }$; then
$$ \begin{align*} \frac { \frac { 1 }{ {y}_{1} }+\frac { 1 }{ {y}_{2} }+…+\frac { 1 }{ {y}_{n} } }{ n }\ge \sqrt [ n ]{ \frac { 1 }{ {y}_{1} }\frac { 1 }{ {y}_{2} }…\frac { 1 }{ {y}_{n} } } \\ \implies \frac { 1 }{ \sqrt [ n ]{ \frac { 1 }{ {y}_{1} }\frac { 1 }{ {y}_{2} }…\frac { 1 }{ {y}_{n} } } }\ge \frac { n }{ \frac { 1 }{ {y}_{1} }+\frac { 1 }{ {y}_{2} }+…+\frac { 1 }{ {y}_{n} } } \\ \implies \sqrt [ n ]{ {y}_{1}{y}_{2}…{y}_{n} }\ge \frac { n }{ \frac { 1 }{ {y}_{1} }+\frac { 1 }{ {y}_{2} }+…+\frac { 1 }{ {y}_{n} } } \end{align*} $$
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