# 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 $x_{1},\ldots,x_{n}>0$; we shall first prove that

 $\sqrt[n]{x_{1}\cdot\ldots\cdot x_{n}}\leq\frac{x_{1}+\ldots+x_{n}}{n}.$

Note that $\log$ is a concave function. Applying it to the arithmetic mean of $x_{1},\ldots,x_{n}$ and using Jensen’s inequality, we see that

 $\displaystyle\log(\frac{x_{1}+\ldots+x_{n}}{n})$ $\displaystyle\geq\frac{\log(x_{1})+\ldots+\log(x_{n})}{n}$ $\displaystyle=\frac{\log(x_{1}\cdot\ldots\cdot x_{n})}{n}$ $\displaystyle=\log{\sqrt[n]{x_{1}\cdot\ldots\cdot x_{n}}}.$

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 ProofOfArithmeticgeometricharmonicMeansInequality 2013-03-22 12:43:07 2013-03-22 12:43:07 mathcam (2727) mathcam (2727) 7 mathcam (2727) Example msc 39B62 msc 26D15 ArithmeticGeometricMeansInequality ProofOfArithmeticGeometricMeansInequalityUsingLagrangeMultipliers