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# 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:

$n(u-v)v^{{n-1}}<u^{n}-v^{n}<n(u-v)u^{{n-1}}$ |

# 2 Second Inequality

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

# 2.1 $-1<x<0$

# 2.2 $0<x$

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

$nx<(1+x)^{n}-1$ |

# 3 Third Inequality

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

# 3.1 $-1<x<0$

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

$nuv^{{n-1}}-nv^{n}<u^{n}-v^{n},$ |

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

$nuv^{{n-1}}-(n-1)v^{n}<u^{n},$ |

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

$n{u\over v}-n+1<\left({u\over v}\right)^{n}.$ |

Taking the reciprocal, we obtain

$\left({v\over u}\right)^{n}<{v\over v+n(u-v)}=1-{n(u-v)\over v+n(u-v)}$ |

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

$(1+x)^{n}-1<{nx\over 1-(n-1)x}.$ |

# 3.2 $0<x<1/(n-1)$

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

$u^{n}-v^{n}<nu^{n}-nu^{{n-1}}v$ |

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

$nu^{{n-1}}v-(n-1)u^{n}<v^{n}$ |

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

$n{v\over u}-n+1<\left({v\over u}\right)^{n}$ |

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

$\left({u\over v}\right)^{n}<{u\over u-n(u-v)}=1+{n(u-v)\over u-n(u-v)}$ |

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

$(1+x)^{n}-1<{nx\over 1-(n-1)x}$ |

and the positivity condition mentioned above becomes $(n-1)x<1$.

## Mathematics Subject Classification

26D99*no label found*

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