proof of inequalities for difference of powers


1 First Inequality

We have the factorization

un-vn=(u-v)k=0n-1ukvn-k-1.

Since the largest term in the sum is is un-1 and the smallest is vn-1, and there are n terms in the sum, we deduce the following inequalitiesMathworldPlanetmath:

n(u-v)vn-1<un-vn<n(u-v)un-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

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

1-(1+x)n<n(-x)

Reversing the signs of both sides yields

nx<(1+x)n-1

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,

nuvn-1-nvn<un-vn,

move the vn to the other side of the inequality,

nuvn-1-(n-1)vn<un,

and divide by vn to obtain

nuv-n+1<(uv)n.

Taking the reciprocal, we obtain

(vu)n<vv+n(u-v)=1-n(u-v)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<nx1-(n-1)x.

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

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

un-vn<nun-nun-1v

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

nun-1v-(n-1)un<vn

and divide by un to obtain

nvu-n+1<(vu)n

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

(uv)n<uu-n(u-v)=1+n(u-v)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<nx1-(n-1)x

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

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