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.

  • $\begingroup$ I think you mean an $(1+x)^n$, not $(1+x)^k$, in your first line. $\endgroup$
    – user71641
    Aug 24, 2013 at 20:20
  • $\begingroup$ Yeah... I'll change it... $\endgroup$ Aug 24, 2013 at 20:20
  • $\begingroup$ $(1+kx)(1+x) = 1 + kx + x + kx^2$. $\endgroup$ Aug 24, 2013 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, 2013 at 20:35
  • $\begingroup$ Also see math.stackexchange.com/q/46562/139123 (a stronger version of the inequality). $\endgroup$
    – David K
    Feb 24, 2015 at 23:43

3 Answers 3


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 $P(1) \implies P(2)$.


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

  • $\begingroup$ This needs a TU! +1 $\endgroup$
    – Amzoti
    Aug 25, 2013 at 0:42

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?

  • $\begingroup$ It implies that the inequality still hold on elimination of $kx^2$. $\endgroup$
    – user112120
    Nov 28, 2013 at 17:04

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