proof of arithmetic-geometric-harmonic means inequality

We can use the Jensen inequality for an easy proof of the arithmetic-geometric-harmonic means inequality.

Let $x_1,\mathrm{\dots },x_n>0$; we shall first prove that

$$\sqrt[n]{x_1\cdot \mathrm{\dots }\cdot x_n}\le \frac{x_1+\mathrm{\dots }+x_n}{n}.$$

Note that $\log$ is a concave function. Applying it to the arithmetic mean of $x_1,\mathrm{\dots },x_n$ and using Jensen’s inequality, we see that

$\log ({\displaystyle \frac{x_1+\mathrm{\dots }+x_n}{n}})$ $\ge {\displaystyle \frac{\log (x_1)+\mathrm{\dots }+\log (x_n)}{n}}$

$={\displaystyle \frac{\log (x_1\cdot \mathrm{\dots }\cdot x_n)}{n}}$

$=\log \sqrt[n]{x_1\cdot \mathrm{\dots }\cdot x_n}.$

Since $\log$ is also a monotone function, it follows that the arithmetic mean is at least as large as the geometric mean.

The proof that the geometric mean is at least as large as the harmonic mean is the usual one (see “proof of arithmetic-geometric-harmonic means inequality”).

Title	proof of arithmetic-geometric-harmonic means inequality
Canonical name	ProofOfArithmeticgeometricharmonicMeansInequality
Date of creation	2013-03-22 12:43:07
Last modified on	2013-03-22 12:43:07
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	7
Author	mathcam (2727)
Entry type	Example
Classification	msc 39B62
Classification	msc 26D15
Related topic	ArithmeticGeometricMeansInequality
Related topic	ProofOfArithmeticGeometricMeansInequalityUsingLagrangeMultipliers