# proof of inequalities for difference of powers

## 1 First Inequality

We have the factorization

$${u}^{n}-{v}^{n}=(u-v)\sum _{k=0}^{n-1}{u}^{k}{v}^{n-k-1}.$$ |

Since the largest term in the sum is is ${u}^{n-1}$ and the smallest is
${v}^{n-1}$, and there are $n$ terms in the sum, we deduce the following
inequalities^{}:

$$ |

## 2 Second Inequality

This inequality is trivial when $x=0$. We split the rest of the proof into two cases.

### 2.1 $$

In this case, we set $u=1$ and $v=1+x$ in the second inequality above:

$$ |

Reversing the signs of both sides yields

$$ |

### 2.2 $$

In this case, we set $u=1+x$ and $v=1$ in the first inequality above:

$$ |

## 3 Third Inequality

This inequality is trivial when $x=0$. We split the rest of the proof into two cases.

### 3.1 $$

Start with the first inequality for differences of powers, expand the left-hand side,

$$ |

move the ${v}^{n}$ to the other side of the inequality,

$$ |

and divide by ${v}^{n}$ to obtain

$$ |

Taking the reciprocal, we obtain

$$ |

Setting $u=1$ and $v=1+x$, and moving a term from one side to the other, this becomes

$$ |

### 3.2 $$

Start with the second inequality for differences of powers, expand the right-hand side,

$$ |

move terms from one side of the inequality to the other,

$$ |

and divide by ${u}^{n}$ to obtain

$$ |

When the left-hand side is positive, (i.e. $nv>(n-1)u$) we can take the reciprocal:

$$ |

Setting $u=1+x$ and $v=1$, and moving a term from one side to the other, this becomes

$$ |

and the positivity condition mentioned above becomes $$.

Title | proof of inequalities for difference of powers |
---|---|

Canonical name | ProofOfInequalitiesForDifferenceOfPowers |

Date of creation | 2013-03-22 15:48:45 |

Last modified on | 2013-03-22 15:48:45 |

Owner | Mathprof (13753) |

Last modified by | Mathprof (13753) |

Numerical id | 14 |

Author | Mathprof (13753) |

Entry type | Proof |

Classification | msc 26D99 |