# proof of arithmetic-geometric-harmonic means inequality

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

 $M=\frac{M+M+M+\cdots+M}{n}\geq\frac{x_{1}+x_{2}+x_{3}+\cdots+x_{n}}{n}$
 $m=\frac{n}{\frac{n}{m}}=\frac{n}{\frac{1}{m}+\frac{1}{m}+\frac{1}{m}+\cdots+% \frac{1}{m}}\leq\frac{n}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+% \cdots+\frac{1}{x_{n}}}$

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

## 1 Case $n=2$

We do first the case $n=2$.

 $\displaystyle(\sqrt{x_{1}}-\sqrt{x_{2}})^{2}$ $\displaystyle\geq$ $\displaystyle 0$ $\displaystyle x_{1}-2\sqrt{x_{1}x_{2}}+x_{2}$ $\displaystyle\geq$ $\displaystyle 0$ $\displaystyle x_{1}+x_{2}$ $\displaystyle\geq$ $\displaystyle 2\sqrt{x_{1}x_{2}}$ $\displaystyle\frac{x_{1}+x_{2}}{2}$ $\displaystyle\geq$ $\displaystyle\sqrt{x_{1}x_{2}}$

## 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.

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

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

 $\displaystyle\frac{x_{1}+x_{2}+\cdots+x_{2^{k}}+x_{2^{k}+1}+\cdots+x_{2^{k+1}}% }{2^{k+1}}$ $\displaystyle\geq$ $\displaystyle\sqrt{\left(\frac{x_{1}+x_{2}+\cdots+x_{2^{k}}}{2^{k}}\right)% \left(\frac{x_{2^{k}+1}+x_{2^{k}+2}+\cdots+x_{2^{k+1}}}{2^{k}}\right)}$ $\displaystyle\geq$ $\displaystyle\sqrt{\sqrt[2^{k}]{x_{1}x_{2}\cdots x_{2^{k}}}\sqrt[2^{k}]{x_{2^{% k}+1}x_{2^{k}+2}\cdots x_{2^{k+1}}}}$

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 $\sqrt[2^{k+1}]{x_{1}x_{2}x_{3}\cdots x_{2^{k+1}}}$ 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

 $\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}\geq\sqrt[n]{x_{1}x_{2}\cdots x_{n}}$

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

 $\displaystyle\frac{x_{1}+x_{2}+\cdots+x_{n-1}+\left(\frac{x_{1}+x_{2}+\cdots+x% _{n-1}}{n-1}\right)}{n}$ $\displaystyle=$ $\displaystyle\frac{(n-1)x_{1}+(n-1)x_{2}+\cdots+(n-1)x_{n-1}+x_{1}+x_{2}+% \cdots+x_{n-1}}{n(n-1)}$ $\displaystyle=$ $\displaystyle\frac{nx_{1}+nx_{2}+\cdots+nx_{n-1}}{n(n-1)}$ $\displaystyle=$ $\displaystyle\frac{x_{1}+x_{2}+\cdots+x_{n-1}}{(n-1)}$

On the other hand

 $\displaystyle\sqrt[n]{x_{1}x_{2}\cdots x_{n-1}\left(\frac{x_{1}+x_{2}+\cdots+x% _{n-1}}{n-1}\right)}$ $\displaystyle=$ $\displaystyle\sqrt[n]{x_{1}x_{2}\cdots x_{n-1}}\sqrt[n]{\frac{x_{1}+x_{2}+% \cdots+x_{n-1}}{n-1}}$

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

 $\left(\frac{x_{1}+x_{2}+\cdots+x_{n-1}}{n-1}\right)^{n}\geq(x_{1}x_{2}\cdots x% _{n})\left(\frac{x_{1}+x_{2}+\cdots+x_{n-1}}{n-1}\right)$

and so

 $\left(\frac{x_{1}+x_{2}+\cdots+x_{n-1}}{n-1}\right)^{n-1}\geq x_{1}x_{2}\cdots x% _{n}$

from where we get that

 $\frac{x_{1}+x_{2}+\cdots+x_{n-1}}{n-1}\geq\sqrt[n-1]{x_{1}x_{2}\cdots x_{n}}.$

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

 $\frac{t_{1}+t_{2}+\cdots+t_{n}}{n}\geq\sqrt[n]{t_{1}t_{2}t_{3}\cdots t_{n}}$

we obtain

 $\frac{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\cdots+\frac{1}{x_{n}}}{% n}\geq\sqrt[n]{\frac{1}{x_{1}}\frac{1}{x_{2}}\frac{1}{x_{3}}\cdots\frac{1}{x_{% n}}}$

and therefore

 $\sqrt[n]{x_{1}x_{2}x_{3}\cdots x_{n}}\geq\frac{n}{\frac{1}{x_{1}}+\frac{1}{x_{% 2}}+\frac{1}{x_{3}}+\cdots+\frac{1}{x_{n}}}$

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