# The inequality $b^n - a^n < (b - a)nb^{n-1}$

I'm trying to figure out why $b^n - a^n < (b - a)nb^{n-1}$.

Using just algebra, we can calculate

$(b - a)(b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1})$

$= (b^n + b^{n-1}a + \ldots + b^{2}a^{n-2} + ba^{n-1}) - (b^{n-1}a + b^{n-2}a^2 + \ldots + ba^{n-1} + a^{n-1})$

$= b^n - a^n,$

but why is it necessarily true that $(b - a)(b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1}) < (b - a)nb^{n-1}$?

Note: I am interested in an answer to that last question, rather than in another way to prove the general inequality in the title...

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For $b>a\ge 0$ it should be obvious, each term is $\le b^{n-1}$. Note things are not quite right, if $b=a$ we do not have $<$. –  André Nicolas Jan 30 '12 at 21:18
Not true if e.g. $b=a$. @Andre: Not true if e.g. $b=0>a$ and $n$ is odd. (Edit: nevermind.) –  anon Jan 30 '12 at 21:24
Yes, I meant for $0 < a < b$. Sorry! –  jamaicanworm Jan 30 '12 at 22:43

This inequality can fail if $a=b$ or $a$ and $b$ have differing signs: e.g. $b=1$, $a=-3$, and $n=3$. So let's assume that $a\not=b$ and $a,b\ge0$. Division yields $$\frac{b^n-a^n}{b-a}=\sum_{k=1}^nb^{n-k}a^{k-1}\tag{1}$$ If $a<b$ then obviously, $\sum\limits_{k=1}^nb^{n-k}a^{k-1}=b^{n-1}\sum\limits_{k=1}^n\left(\frac{a}{b}\right)^{k-1}< nb^{n-1}$, thus $\dfrac{b^n-a^n}{b-a}< nb^{n-1}$ and because $b-a>0$, $$b^n-a^n<(b-a)nb^{n-1}\tag{2}$$ If $a>b$ then obviously, $\sum\limits_{k=1}^nb^{n-k}a^{k-1}=b^{n-1}\sum\limits_{k=1}^n\left(\frac{a}{b}\right)^{k-1}> nb^{n-1}$, thus $\dfrac{b^n-a^n}{b-a}> nb^{n-1}$ and because $b-a<0$, $$b^n-a^n<(b-a)nb^{n-1}\tag{3}$$ If both $a$ and $b$ are negative, the inequality holds for even $n$, and is reversed for odd $n$.

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You ask why it is necessarily true that $$(b - a)(b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1}) < (b - a)nb^{n-1}.$$ A quick answer is that it is not necessarily true. We cannot have strict inequality if $b=a$. And there are other issues. For example, if $n=1$, we always have equality. And the inequality sometimes reverses when one of $a$ or $b$ is negative.

We deal in detail with the cases where $a$ and $b$ are both $\ge 0$. Suppose first that $b>a$, and let $n>1$. Then we want to show that $$b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1} < nb^{n-1}.$$ The term $b^{n-2}a$ is strictly less than $b^{n-1}$, and there is such a term, since $n>1$. The remaining terms (if any) are also strictly less than $b^{n-1}$. There is a total of $n$ terms, so their sum is strictly less than $nb^{n-1}$.

A similar argument deals with $b<a$. Recall that inequality is reversed when we divide both sides by a negative number. So we want to show that $$b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1} > nb^{n-1}.$$ Since $b<a$, the term $a^{n-1}$ is strictly bigger than $b^{n-1}$. We don't even need $n>1$. All the other terms are $\ge b^{n-1}$. There is a total of $n$ terms, and the inequality follows.

We stop at this point. The work is incomplete, since we have not dealt with negative numbers. Let $n=3$, $b=0$, and $a=-1$. Then $(b-a)(b^2+ba+a^2)$ is positive, but $(b-a)(3)b^2$ is $0$, so the inequality does not hold. With some work, one can track down all the situations where the inequality does hold.

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