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


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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.
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Classification:
AMS MSC26D15 (Real functions :: Inequalities :: Inequalities for sums, series and integrals)

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