PlanetMath (more info)
 Math for the people, by the people.
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
sections
Encyclopædia
Papers
Books
Expositions

meta
Requests (232)
Orphanage
Unclass'd
Unproven (519)
Corrections (22)
Classification

talkback
Polls
Forums
Feedback
Bug Reports

downloads
Snapshots
PM Book

information
News
Docs
Wiki
ChangeLog
TODO List
Legalese
About
Owner confidence rating: High Entry average rating: No information on entry rating
[parent] proof of arithmetic-geometric-harmonic means inequality (Proof)

Let $M$ be $\max\{x_1,x_2,x_3,\ldots,x_n\}$ and let $m$ be $\min\{x_1,x_2,x_3,\ldots,x_n\}$.

Then

\begin{displaymath}M=\frac{M+M+M+\cdots+M}{n}\geq\frac{x_1+x_2+x_3+\cdots+x_n}{n}\end{displaymath}


\begin{displaymath}m=\frac{n}{\frac{n}{m}}= \frac{n}{\frac{1}{m}+\frac{1}{m}+\fr... ...\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\cdots+\frac{1}{x_n}}\end{displaymath}

where all the summations have $n$ terms. So we have proved in this way the two inequalities at the extremes.

Now we shall prove the inequality between arithmetic mean and geometric mean.

Case $n=2$

We do first the case $n=2$.
\begin{eqnarray*} (\sqrt{x_1}-\sqrt{x_2})^2 &\geq& 0\ x_1-2\sqrt{x_1x_2}+x_2&\... ...+x_2&\geq&2\sqrt{x_1x_2}\ \frac{x_1+x_2}{2}&\geq&\sqrt{x_1x_2} \end{eqnarray*}


Case $n=2^k$

Now we prove the inequality for any power of $2$ (that is, $n=2^k$ for some integer $k$) by using mathematical induction.
\begin{eqnarray*} \lefteqn{ \frac{x_1+x_2+\cdots+x_{2^k}+x_{2^k+1}+\cdots+x_{2^{... ... \frac{x_{2^k+1}+x_{2^k+2}+\cdots+x_{2^{k+1}}}{2^k} \right)} {2} \end{eqnarray*}


and using the case $n=2$ on the last expression we can state the following inequality
\begin{eqnarray*} \lefteqn{ \frac{x_1+x_2+\cdots+x_{2^k}+x_{2^k+1}+\cdots+x_{2^{... ...dots x_{2^k}} \sqrt[2^k]{x_{2^k+1}x_{2^k+2}\cdots x_{2^{k+1}}} } \end{eqnarray*}


where the last inequality was obtained by applying the induction hypothesis with $n=2^k$. Finally, we see that the last expression is equal to $\sqrt[2^{k+1}]{x_1x_2x_3\cdots x_{2^{k+1}}}$ and so we have proved the truth of the inequality when the number of terms is a power of two.

Inequality for $n$ numbers implies inequality for $n-1$

Finally, we prove that if the inequality holds for any $n$, it must also hold for $n-1$, and this proposition, combined with the preceding proof for powers of $2$, is enough to prove the inequality for any positive integer.

Suppose that

\begin{displaymath}\frac{x_1+x_2+\cdots+x_n}{n}\geq\sqrt[n]{x_1x_2\cdots x_n}\end{displaymath}

is known for a given value of $n$ (we just proved that it is true for powers of two, as example). Then we can replace $x_n$ with the average of the first $n-1$ numbers. So
\begin{eqnarray*} \lefteqn{ \frac{ x_1+x_2+\cdots+x_{n-1}+ \left(\frac{x_1+x_2+\... ...x_{n-1}}{n(n-1)}\ &=& \frac{x_1+x_2 + \cdots + x_{n-1}}{(n-1)} \end{eqnarray*}


On the other hand

\begin{eqnarray*} \lefteqn{\sqrt[n]{ x_1x_2\cdots x_{n-1} \left(\frac{x_1+x_2+\c... ...1x_2\cdots x_{n-1}} \sqrt[n]{\frac{x_1+x_2+\cdots+x_{n-1}}{n-1}} \end{eqnarray*}


which, by hypothesis (the inequality holding for $n$ numbers) and the observations made above, leads to:
\begin{displaymath}\left(\frac{x_{1}+x_{2}+\cdots+x_{n-1}}{n-1}\right)^{n}\ge (x... ...2}\cdots x_{n})\left(\frac{x_1+x_2+\cdots+x_{n-1}}{n-1}\right) \end{displaymath}

and so
\begin{displaymath} \left(\frac{x_{1}+x_{2}+\cdots+x_{n-1}}{n-1}\right)^{n-1}\ge x_{1}x_{2}\cdots x_{n} \end{displaymath}

from where we get that
\begin{displaymath} \frac{x_{1}+x_{2}+\cdots+x_{n-1}}{n-1}\ge\sqrt[n-1]{x_{1}x_{2}\cdots x_{n}}.\end{displaymath}

So far we have proved the inequality between the arithmetic mean and the geometric mean. The geometric-harmonic inequality is easier. Let $t_i$ be $1/x_{i}$.

From

\begin{displaymath}\frac{t_{1}+t_{2}+\cdots+t_{n}}{n}\geq\sqrt[n]{t_{1}t_{2}t_{3}\cdots t_{n}}\end{displaymath}

we obtain
\begin{displaymath}\frac{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\cdots+... ...c{1}{x_{1}}\frac{1}{x_{2}}\frac{1}{x_{3}}\cdots\frac{1}{x_{n}}}\end{displaymath}

and therefore
\begin{displaymath}\sqrt[n]{x_{1}x_{2}x_{3}\cdots x_{n}}\geq \frac{n}{ \frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\cdots+\frac{1}{x_{n}} }\end{displaymath}

and so, our proof is completed.



"proof of arithmetic-geometric-harmonic means inequality" is owned by drini. [ owner history (1) ]
(view preamble | get metadata)

View style:

See Also: arithmetic mean, geometric mean, harmonic mean, general means inequality, weighted power mean, power mean


This object's parent.
Log in to rate this entry.
(view current ratings)

Cross-references: hypothesis, positive, proof, proposition, power of two, number, induction hypothesis, expression, induction, integer, power, geometric mean, arithmetic mean, inequalities, terms, summations

This is version 3 of proof of arithmetic-geometric-harmonic means inequality, born on 2002-05-30, modified 2005-09-29.
Object id is 2970, canonical name is ProofOfArithmeticGeometricHarmonicMeansInequality.
Accessed 13378 times total.

Classification:
AMS MSC: 26D15 (Real functions :: Inequalities :: Inequalities for sums, series and integrals)
Pending Errata and Addenda
None.
[ View all 2 ]
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | add example | add (any)