19
$\begingroup$

I'm asked to used induction to prove Bernoulli's Inequality: If $1+x>0$, then $(1+x)^n\geq 1+nx$ for all $n\in\mathbb{N}$. This what I have so far:

Let $n=1$. Then $1+x\geq 1+x$. This is true. Now assume that the proposed inequality holds for some arbitrary $k$, namely that $$1+x>0\implies (1+x)^k\geq 1+kx,~\forall~k\in\mathbb{N}\setminus\{1\}$$ is true. We want to show that the proposed inequality holds for $k+1$. Thus multiplication of $(1+x)$ on each side of the above inequality produces the following result: $$(1+x)(1+x)^k\geq (1+kx)(1+x)\implies (1+x)^{k+1}\geq 1+x+kx+kx^2\cdots\cdots\cdots$$

I'm not sure where to go from here.

$\endgroup$
  • $\begingroup$ I think you mean an $(1+x)^n$, not $(1+x)^k$, in your first line. $\endgroup$ – user71641 Aug 24 '13 at 20:20
  • $\begingroup$ Yeah... I'll change it... $\endgroup$ – Loie Benedicte Aug 24 '13 at 20:20
  • $\begingroup$ $(1+kx)(1+x) = 1 + kx + x + kx^2$. $\endgroup$ – Daniel Fischer Aug 24 '13 at 20:25
  • $\begingroup$ I think you want to change your induction hypothesis slightly; you want to assume that $(1+x)^k\ge1+kx$ for some $k$ in $N$. (Notice this is not for all k, and we are allowing $k=1$.) $\endgroup$ – user84413 Aug 24 '13 at 20:35
  • $\begingroup$ Also see math.stackexchange.com/q/46562/139123 (a stronger version of the inequality). $\endgroup$ – David K Feb 24 '15 at 23:43
10
$\begingroup$

You're almost done. $(1+x)^{k+1}\geq (1+kx)(1+x)=1+kx^2+(k+1)x\geq1+(k+1)x$ since $kx^2\geq 0$.

Also in your assumption, "$1+x>0\implies (1+x)^k\geq 1+kx,~\forall~k\in\mathbb{N}\setminus\{1\}$", you shouldn't write $\forall~k\in\mathbb{N}\setminus\{1\}$. Because if you assume this is true for all $k\in \mathbb{N}$, then you've already assumed it is true for $k+1$. Also, you shouldn't let $k\ne 1$, since then your argument doesn't allow you to conclude $P(1) \implies P(2)$.

$\endgroup$
9
$\begingroup$

$$\begin{align} (1 + x)^{k+1} & \geq (1+kx)(1+x) \\ \\ & = 1 + kx + x + kx^2 \\ \\ & = 1 + (k+1)x + kx^2 \\ \\ & \geq 1+ (k+1) x\end{align}$$ as desired.

Remark: The inductive hypothesis $P(k)$ is assumed only for some arbitrary $k \in \mathbb N$; i.e., the point of an inductive proof is to then show that it is indeed true for all $n \in \mathbb N$, by first showing that $P(1) \land (P(k) \implies P(k+1)).$

$\endgroup$
  • $\begingroup$ This needs a TU! +1 $\endgroup$ – Amzoti Aug 25 '13 at 0:42
2
$\begingroup$

You have $(1 + x)^{k+1} \geq 1 + (k+1)x + kx^2$. Keeping in mind that $kx^2 \geq 0$, what does this imply about your inequality?

$\endgroup$
  • $\begingroup$ It implies that the inequality still hold on elimination of $kx^2$. $\endgroup$ – user112120 Nov 28 '13 at 17:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.