# proof of arithmetic-geometric-harmonic means inequality

For the Arithmetic Geometric Inequality, I claim it is enough to prove that if
${\prod}_{i=1}^{n}{x}_{i}=1$ with ${x}_{i}\ge 0$ then ${\sum}_{i=1}^{n}{x}_{i}\ge n$. The arithmetic geometric inequality for ${y}_{1},\mathrm{\dots},{y}_{n}$ will follow by taking
${x}_{i}=\frac{{y}_{i}}{\sqrt[n]{{\prod}_{k=1}^{n}{y}_{k}}}$. The geometric harmonic inequality^{} follows from the arithmetic geometric by taking ${x}_{i}=\frac{1}{{y}_{i}}$.

So, we show that if ${\prod}_{i=1}^{n}{x}_{i}=1$ with ${x}_{i}\ge 0$ then ${\sum}_{i=1}^{n}{x}_{i}\ge n$ by induction^{} on $n$.

Clear for $n=1$.

Induction Step: By reordering indices we may assume the ${x}_{i}$ are increasing, so ${x}_{n}\ge 1\ge {x}_{1}$. Assuming the statement is true for $n-1$, we have ${x}_{2}+\mathrm{\cdots}+{x}_{n-1}+{x}_{1}{x}_{n}\ge n-1$. Then,

$$\sum _{i=1}^{n}{x}_{i}\ge n-1+{x}_{n}+{x}_{1}-{x}_{1}{x}_{n}$$ |

by adding ${x}_{1}+{x}_{n}$ to both sides and subtracting ${x}_{1}{x}_{n}$. And so,

$\sum _{i=1}^{n}}{x}_{i$ | $\ge n+({x}_{n}-1)+({x}_{1}-{x}_{1}{x}_{n})$ | ||

$=n+({x}_{n}-1)-{x}_{1}({x}_{n}-1)$ | |||

$=n+({x}_{n}-1)(1-{x}_{1})$ | |||

$\ge n$ |

The last line follows since ${x}_{n}\ge 1\ge {x}_{1}$.

Title | proof of arithmetic-geometric-harmonic means inequality |
---|---|

Canonical name | ProofOfArithmeticgeometricharmonicMeansInequality |

Date of creation | 2013-03-22 15:09:37 |

Last modified on | 2013-03-22 15:09:37 |

Owner | Mathprof (13753) |

Last modified by | Mathprof (13753) |

Numerical id | 10 |

Author | Mathprof (13753) |

Entry type | Proof |

Classification | msc 26D15 |