3

I think I understand how induction works, but I wasn't able to justify all the steps necessary to prove this proposition: $(1+x)^n≥1+nx, ∀x>-1, ∀n∈N$

One thing that confuses me is that I don't know whether I should use induction with both x and n. I didn't pay attention to the x and I still couldn't justify all the steps.

Thanks.

Kim Jong Un
  • 14,794
  • 1
  • 24
  • 49
Rrmm
  • 163

5 Answers5

3

The base case $n=0$ is just $1\geq 1$ which is true. For the induction step, note that $$ (1+x)^{n+1}=(1+x)(1+x)^n\\ \geq (1+x)(1+nx)=1+x(n+1)+nx^2\geq1+x(n+1). $$ Note that in going from the first line to the second, we need both the induction hypothesis $(1+x)^n\geq 1+nx$ and $x>-1$.

yurnero
  • 10,505
2

We have \begin{align*} (1+x)^n-1&=(1+x-1)\left[1+(1+x)+\cdots+(1+x)^{n-1}\right]\\&=x\left[1+(1+x)+\cdots+(1+x)^{n-1}\right] \end{align*} Easy to find that $[1+(1+x)+\cdots+(1+x)^{n-1}]\geqslant{n}$ while $x\geqslant0$

if$-1<x<0,[1+(1+x)+\cdots+(1+x)^{n-1}]<n$, Since $x<0,(1+x)^n-1>nx.$ $viz. (1+x)^n-1\geqslant{n}x$, more generally ,we find that the inequality is still valid for $-2\leqslant{x}\leqslant-1$.

Shine Mic
  • 835
  • 4
  • 16
Faye Tao
  • 1,814
1

Consider the concave function $f(z)=\log(1+z)$ defined on $(-1,\infty)$. An alternative approach to induction involves the Jensen's inequality: noting $f(0)=0$, we have: $$ \frac{1}{n}\log(1+nx)=\frac{1}{n}f(nx)+\frac{n-1}{n}f(0)\leq f\left(\frac{nx}{n}+\frac{(n-1)0}{n}\right)=f(x)=\log(1+x). $$ Thus, $$ \frac{1}{n}\log(1+nx)\leq\log(1+x)\iff\log(1+nx)\leq\log[(1+x)^n]\iff 1+nx\leq(1+x)^n. $$

Kim Jong Un
  • 14,794
  • 1
  • 24
  • 49
1

Fix $x\in\mathbb{R}$ with $x>-1$ and $x\neq 0$. For each $n\geq 2$, let $S(n)$ be the statement that $(1+x)^n>1+nx$ holds.

Base step: Since $x\neq 0, x^2>0$, and so $(1+x)^2=1+2x+x^2>1+2x, S(2)$ holds.

Inductive step: Fix $k\geq 2$, and suppose that $S(k)$ holds; that is, $(1+x)^k>1+kx$. Before proving $S(k+1)$, note the following "subtlety" concerning inequalities: If $a>b$ and $c>0$, then $ac>bc$; if $c<0$, then $a<b$ implies $ac<bc$. Finally, the proof of $S(k+1)$: \begin{align} (1+x)^{k+1} &= (1+x)^k(1+x)\\[0.5em] &> (1+kx)(1+x)\tag{since $x+1>0$}\\[0.5em] &= 1+x+kx+kx^2\\[0.5em] &= 1+(k+1)x+kx^2\\[0.5em] &> 1+(k+1)x.\tag{since $x\neq 0$} \end{align} Thus, $S(k+1)$ is true, completing the inductive step.

By mathematical induction, for all $n\geq 2$, the statement $S(n)$ is true. $\blacksquare$

Daniel W. Farlow
  • 22,531
0

If you want to procced by induction, at first step you have to define a inductive set:

Let be $X\subset\mathbb{N}$ such: $$X=\{n\in\mathbb{N}:\;\;(1+x)^n\geq1+nx,\,\forall x\geq-1\}$$ $1\in X$, because for $ x\geq-1$ we have: $$(1+x)^1=(1+1.x)\geq (1+1.x) $$ Suppose $n\in \mathbb{N}$, then
$$(1+x)^{n+1}=(1+x)(1+x)^n\displaystyle\geq^{HI}(1+x)(1+nx)=1+(n+1)x+nx^2≥1+(n+1)x$$ Then, $n\in X \Rightarrow s(n)\in X,$ for $P3$ (Third axiom of Peano) $X=\mathbb{N}$ so the inequation holds for all $n\in\mathbb{N}$ and $x\geq-1 \in\mathbb{R}$.

L F
  • 3,644