# 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 $\mathrm{log}$ 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

$\mathrm{log}({\displaystyle \frac{{x}_{1}+\mathrm{\dots}+{x}_{n}}{n}})$ | $\ge {\displaystyle \frac{\mathrm{log}({x}_{1})+\mathrm{\dots}+\mathrm{log}({x}_{n})}{n}}$ | ||

$={\displaystyle \frac{\mathrm{log}({x}_{1}\cdot \mathrm{\dots}\cdot {x}_{n})}{n}}$ | |||

$=\mathrm{log}\sqrt[n]{{x}_{1}\cdot \mathrm{\dots}\cdot {x}_{n}}.$ |

Since $\mathrm{log}$ 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 |