proof of arithmetic-geometric-harmonic means inequality

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

Then

$$M=\frac{M+M+M+\mathrm{\cdots }+M}{n}\ge \frac{x_1+x_2+x_3+\mathrm{\cdots }+x_n}{n}$$

$$m=\frac{n}{\frac{n}{m}}=\frac{n}{\frac{1}{m}+\frac{1}{m}+\frac{1}{m}+\mathrm{\cdots }+\frac{1}{m}}\le \frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\mathrm{\cdots }+\frac{1}{x_n}}$$

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.

1 Case $n=2$

We do first the case $n=2$.

${(\sqrt{x_1}-\sqrt{x_2})}^2$ $\ge$ $0$

$x_1-2\sqrt{x_1{x}_2}+x_2$ $\ge$ $0$

$x_1+x_2$ $\ge$ $2\sqrt{x_1{x}_2}$

${\displaystyle \frac{x_1+x_2}{2}}$ $\ge$ $\sqrt{x_1{x}_2}$

2 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.

${\displaystyle \frac{x_1+x_2+\mathrm{\cdots }+x_{2^k}+x_{2^k+1}+\mathrm{\cdots }+x_{2^{k+1}}}{2^{k+1}}}$

$=$ ${\displaystyle \frac{\left(\frac{x_1+x_2+\mathrm{\cdots }+x_{2^k}}{2^k}\right)+\left(\frac{x_{2^k+1}+x_{2^k+2}+\mathrm{\cdots }+x_{2^{k+1}}}{2^k}\right)}{2}}$

and using the case $n=2$ on the last expression we can state the following inequality

${\displaystyle \frac{x_1+x_2+\mathrm{\cdots }+x_{2^k}+x_{2^k+1}+\mathrm{\cdots }+x_{2^{k+1}}}{2^{k+1}}}$

$\ge$ $\sqrt{\left({\displaystyle \frac{x_1+x_2+\mathrm{\cdots }+x_{2^k}}{2^k}}\right)\left({\displaystyle \frac{x_{2^k+1}+x_{2^k+2}+\mathrm{\cdots }+x_{2^{k+1}}}{2^k}}\right)}$

$\ge$ $\sqrt{\sqrt[2^k]{x_1{x}_2\mathrm{\cdots }{x}_{2^k}}\sqrt[2^k]{x_{2^k+1}{x}_{2^k+2}\mathrm{\cdots }{x}_{2^{k+1}}}}$

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 $\sqrt[2^{k+1}]{x_1{x}_2{x}_3\mathrm{\cdots }{x}_{2^{k+1}}}$ and so we have proved the truth of the inequality when the number of terms is a power of two.

3 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

$$\frac{x_1+x_2+\mathrm{\cdots }+x_n}{n}\ge \sqrt[n]{x_1{x}_2\mathrm{\cdots }{x}_n}$$

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

${\displaystyle \frac{x_1+x_2+\mathrm{\cdots }+x_{n-1}+\left(\frac{x_1+x_2+\mathrm{\cdots }+x_{n-1}}{n-1}\right)}{n}}$

$=$ ${\displaystyle \frac{(n-1)x_1+(n-1)x_2+\mathrm{\cdots }+(n-1)x_{n-1}+x_1+x_2+\mathrm{\cdots }+x_{n-1}}{n(n-1)}}$

$=$ ${\displaystyle \frac{n{x}_1+n{x}_2+\mathrm{\cdots }+n{x}_{n-1}}{n(n-1)}}$

$=$ ${\displaystyle \frac{x_1+x_2+\mathrm{\cdots }+x_{n-1}}{(n-1)}}$

On the other hand

$\sqrt[n]{x_1{x}_2\mathrm{\cdots }{x}_{n-1}\left({\displaystyle \frac{x_1+x_2+\mathrm{\cdots }+x_{n-1}}{n-1}}\right)}$

$=$ $\sqrt[n]{x_1{x}_2\mathrm{\cdots }{x}_{n-1}}\sqrt[n]{{\displaystyle \frac{x_1+x_2+\mathrm{\cdots }+x_{n-1}}{n-1}}}$

which, by hypothesis (the inequality holding for $n$ numbers) and the observations made above, leads to:

$${\left(\frac{x_1+x_2+\mathrm{\cdots }+x_{n-1}}{n-1}\right)}^n\ge (x_1{x}_2\mathrm{\cdots }{x}_n)\left(\frac{x_1+x_2+\mathrm{\cdots }+x_{n-1}}{n-1}\right)$$

and so

$${\left(\frac{x_1+x_2+\mathrm{\cdots }+x_{n-1}}{n-1}\right)}^{n-1}\ge x_1{x}_2\mathrm{\cdots }{x}_n$$

from where we get that

$$\frac{x_1+x_2+\mathrm{\cdots }+x_{n-1}}{n-1}\ge \sqrt[n-1]{x_1{x}_2\mathrm{\cdots }{x}_n}.$$

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

$$\frac{t_1+t_2+\mathrm{\cdots }+t_n}{n}\ge \sqrt[n]{t_1{t}_2{t}_3\mathrm{\cdots }{t}_n}$$

we obtain

$$\frac{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\mathrm{\cdots }+\frac{1}{x_n}}{n}\ge \sqrt[n]{\frac{1}{x_1}\frac{1}{x_2}\frac{1}{x_3}\mathrm{\cdots }\frac{1}{x_n}}$$

and therefore

$$\sqrt[n]{x_1{x}_2{x}_3\mathrm{\cdots }{x}_n}\ge \frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\mathrm{\cdots }+\frac{1}{x_n}}$$

and so, our proof is completed.

Title	proof of arithmetic-geometric-harmonic means inequality
Canonical name	ProofOfArithmeticgeometricharmonicMeansInequality
Date of creation	2013-03-22 12:41:25
Last modified on	2013-03-22 12:41:25
Owner	drini (3)
Last modified by	drini (3)
Numerical id	6
Author	drini (3)
Entry type	Proof
Classification	msc 26D15
Related topic	ArithmeticMean
Related topic	GeometricMean
Related topic	HarmonicMean
Related topic	GeneralMeansInequality
Related topic	WeightedPowerMean
Related topic	PowerMean