I am working on getting the hang of proofs by induction, and I was hoping the community could give me feedback on how to format a proof of this nature:

Let $x > -1$ and $n$ be a positive integer. Prove Bernoulli's inequality: $$ (1+x)^n \ge 1+nx$$


Base Case: For $n=1$, $1+x = 1+x$ so the inequality holds.

Induction Assumption: Assume that for some integer $k\ge1$, $(1+x)^k \ge 1+kx$.

Inductive Step: We must show that $(1+x)^{k+1} \ge 1+(k+1)x$

Proof of Inductive Step: $$\begin{align*} (1+x)^k &\ge 1+kx \\ (1+x)(1+x)^k &\ge (1+x)(1+kx)\\ (1+x)^{k+1} &\ge 1 + (k+1)x + kx^2 \\ 1 + (k+1)x + kx^2 &> 1+(k+1)x \quad (kx^2 >0) \\ \Rightarrow (1+x)^{k+1} &\ge 1 + (k+1)x \qquad \qquad \qquad \square \end{align*}$$

  • $\begingroup$ Where did you uses $x>-1$? Hint: you did use it. $\endgroup$ Aug 12 '12 at 15:16
  • $\begingroup$ When I claimed that $kx^2 >0$ $\endgroup$
    – Moderat
    Aug 12 '12 at 15:18
  • 2
    $\begingroup$ Actually, you need it when you multiply both sides by $1+x$. Also, since $x$ can be 0, $kx^2\ge0$ $\endgroup$
    – Mike
    Aug 12 '12 at 16:06
  • $\begingroup$ related: math.stackexchange.com/q/475309/173147 $\endgroup$
    – glS
    Feb 24 '15 at 23:54
  • $\begingroup$ Why we need to multiply both sides by $1+x$ in the second row of inductive step? $\endgroup$
    – Andrej
    Nov 11 '17 at 14:16

What you have is perfectly acceptable. The calculations could be organized a little more neatly:

$$\begin{align*} (1+x)^{k+1}&=(1+x)(1+x)^k\\ &\ge(1+x)(1+kx)\\ &=1+(k+1)x+kx^2\\ &\ge1+(k+1)x\;, \end{align*}$$

since $kx^2\ge 0$. This completes the induction step.


This looks fine to me. Just a small note on formatting of the inequalities: I would combine the third and fourth inequalities as $$ (1+x)^{k+1} \geq 1+(k+1)x+kx^2>1+(k+1)x, $$ so there is no need of the fifth line. Or even $$ (1+x)^{k+1} = (1+x)(1+x)^{k} \geq (1+x)(1+kx)=1+(k+1)x+kx^2>1+(k+1)x. $$


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