# 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}

## 2 Second Inequality

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

### 2.1 $-1

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

 $1-(1+x)^{n}

Reversing the signs of both sides yields

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

### 2.2 $0

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

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

 $nuv^{n-1}-nv^{n}

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

 $nuv^{n-1}-(n-1)v^{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

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

 $u^{n}-v^{n}

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

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

Title proof of inequalities for difference of powers ProofOfInequalitiesForDifferenceOfPowers 2013-03-22 15:48:45 2013-03-22 15:48:45 Mathprof (13753) Mathprof (13753) 14 Mathprof (13753) Proof msc 26D99