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 followinginequalities:

2 Second Inequality

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

2.1

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

Reversing the signs of both sides yields

2.2

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

3 Third Inequality

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

3.1

Start with the first inequality for differences of powers, expandthe 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 sideto the other, this becomes

3.2

Start with the second inequality for differences of powers, expandthe 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 sideto the other, this becomes

and the positivity condition mentioned above becomes .