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I have already figured out a simple proof by induction for this problem. Is there any other way to do it?

Prove $b^n-1 \geq n(b-1)$. For $b>1$ and $n \geq 0$.

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    $\begingroup$ Well what is your proof? $\endgroup$ Sep 17, 2015 at 21:09
  • $\begingroup$ I don't think you can use induction because $n$ is not discrete. $\endgroup$
    – 1-___-
    Sep 17, 2015 at 21:12
  • $\begingroup$ @user2770287: $n$ is not a natural number? $\endgroup$
    – Bernard
    Sep 17, 2015 at 21:14
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    $\begingroup$ Your title is confusing: is Walter Rudin an interesting inequality? $\endgroup$ Sep 17, 2015 at 21:28
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    $\begingroup$ take $b=4, n =0.5$ then the conclusion is not true $\endgroup$ Sep 17, 2015 at 21:35

9 Answers 9

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$$b^n-1=(b-1)(b^{n-1}+\cdots+b+1)\ge(b-1)(1+\cdots+1)=n(b-1).$$

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  • $\begingroup$ you arenot right use Geometric Series for $\frac{1}{b}$ $\endgroup$
    – R.N
    Sep 17, 2015 at 21:15
  • $\begingroup$ For natural $n$, lovely proof. $\endgroup$
    – Macavity
    Sep 18, 2015 at 5:42
  • $\begingroup$ It should be pointed out that this is an exercise from Principles of Mathematical Analysis. $n\in\Bbb N_+$ is a hypothesis. :) The exercise is from Chapter 1 before derivatives and integrals are defined. $\endgroup$ Sep 18, 2015 at 6:24
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Take $f(x) = x^n$, then mean value theorem says that there exists $c \in (1,b)$ such that $f(b) - f(1) = f'(c)(b-1)$, i.e. $$b^n - 1 = nc^{n-1}(b-1)$$ thus $b^n - 1 > n(b-1)$ since $c >1$. Actually we can see that one needs $n \geq 1$(or $n=0$) for the conclusion to be true

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Actually the inequality is true for all $\color{red}{b>0}$.

The initialisation is trivial.

Inductive step:

Suppose $b^n-1>n(b-1)$ for some $n>1$. Rewrite this as $b^n>1+n(b-1)$ and multiply both sides by $b=b-1+1$, which is positive. We get: \begin{align*}b^{n+1}&>1+b-1+n(b-1+1)(b-1)=1+(n+1)(b-1)+n(b-1)^2\\&>1+(n+1)(b-1).\end{align*}

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  • $\begingroup$ The first equality on the last line is false... $\endgroup$ Sep 17, 2015 at 23:51
  • $\begingroup$ You're right. However, I'm sure of the result: $b>0$ is enough. $\endgroup$
    – Bernard
    Sep 17, 2015 at 23:54
  • $\begingroup$ Yes, it is true tha for all $b$, $b(b-1)\geq b-1$ $\endgroup$ Sep 17, 2015 at 23:55
  • $\begingroup$ OK, I've found the typo. Thanks! $\endgroup$
    – Bernard
    Sep 17, 2015 at 23:56
  • $\begingroup$ Nope. Still wrong. Now the last equality is wrong. You need to get rid of the $b^2$, which requires an additional inequality $\endgroup$ Sep 17, 2015 at 23:58
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MVT: Consider $f(x) = x^n \Rightarrow f'(x) = nx^{n-1} \geq n, x \geq 1 \Rightarrow f(b) - f(1) = f'(c)(b-1) \Rightarrow b^n-1 \geq n(b-1)$

Bernoulli: Take $b = 1+a, a > 0 \Rightarrow b^n - 1 = (1+a)^n - 1 \geq 1+an-1 = an = n(b-1)$.

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  • $\begingroup$ Awesome Solution $\endgroup$
    – user249332
    Sep 17, 2015 at 21:36
  • $\begingroup$ Note, Bernoulli needs $n \ge 1$... $\endgroup$
    – Macavity
    Sep 18, 2015 at 5:05
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One way: $b^n-1=(b-1)(1+b+...+b^{n-1})>(b-1)()1+1+...+1) =n(b-1)$

Other way: consider function $f(b)=b^n-1-n(b-1)$, for $b>1$. Taking derivative and conculde that it's increasing on $b \in (1,\infty)$. Thus $f(b)>f(0)=0$.

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Yes, there is another answer: we have that $$(b - 1) ( b^{n - 1} + b^{n - 2} + \cdots + b + 1) = b^{n} - 1.$$ This means that we can write $$ \frac{b^n - 1}{b - 1} = b^{n - 1} + b^{n - 2} + \cdots + b + 1.$$ The right hand side has $n$ terms all of which (by hypothesis on $b$) are greater than or equal to 1; thus we can conclude that $$ \frac{b^n - 1}{b - 1} \ge n.$$

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  • $\begingroup$ Your answer is the same as Tim Raczkowski's. $\endgroup$ Sep 17, 2015 at 23:40
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Here is yet another way...

Case 1: $b, n \ge 1$

$$x > 1\implies x^{n-1} \ge 1 \implies \int_1^b x^{n-1} dx \ge \int_1^b 1 dx \implies b^n-1 \ge n(b-1)$$

Case 2: $b \in (0, 1)$ and $n \ge 1$.

$$x < 1\implies x^{n-1} \le 1 \implies \int_b^1 x^{n-1} dx \le \int_b^1 1 dx \implies b^n-1 \ge n(b-1)$$

Note however that if $n \in (0, 1)$ your inequality does not hold, rather the reverse does. Same proof...

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I had also noticed @PetiteEtincelle's counterexample, $(b,n) = (4,\frac{1}{2})$. And this suggested an intended fixup and proof...

The stated inequality holds for $b,n \in [1,\infty) \subset \mathbb{R}$.

True by inspection at $b=1,n=1$. In fact, true by inspection at $b=1, n \in [1,\infty)$. Partial differentiation with respect to $b$ gives $n b^{n-1} \geq n$ which is evidently strictly true for $b>1, n>1$. So for any point $(b,n) \in [1,\infty) \times [1,\infty)$, we can start at $(1,n)$, where the stated inequality holds, then increase the first coordinate to $b$, and the inequality will hold at every point along the path (because our derivative shows the "big side" grows strictly faster along this path than the "small side").

BTW: Before anyone "complains"... Yes, I know that the notation "$n$" suggests that $n \in \mathbb{Z}$ or $\mathbb{N}$. But "$b$" doesn't suggest anything, so using "obvious" default domains isn't so obvious.

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Proof By Induction:
Inductive Hypothesis: $b^{n}-1\geq n(b-1)$
Base: $n=0$. $0=0$ True.
Inductive Step:
$b^{n+1}-1\geq (n+1)(b-1)$
$b^{n+1}-1\geq n(b-1)+b-1$
$b^{n+1}\geq n(b-1)+b$
$b*b^{n}\geq b*[n(b-1)+1]$
$b*[n(b-1)+1]\geq n(b-1)+b$ $\implies b\geq 1$, Hypothesis. QED

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