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: $$ (1+x)^n \ge 1+nx$$
Proof:
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: $$\begin{align*} (1+x)^k &\ge 1+kx \\ (1+x)(1+x)^k &\ge (1+x)(1+kx)\\ (1+x)^{k+1} &\ge 1 + (k+1)x + kx^2 \\ 1 + (k+1)x + kx^2 &> 1+(k+1)x \quad (kx^2 >0) \\ \Rightarrow (1+x)^{k+1} &\ge 1 + (k+1)x \qquad \qquad \qquad \square \end{align*}$$
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$\begingroup$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.
$\endgroup$ $\begingroup$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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