I'm having trouble understanding a step of induction.

Question

The problem my teacher presented was to prove, $(1 + x)^n \geq 1 + nx$ for all real numbers $x > -1$ and integers $n \geq 2$. The way it was done in class is:

1. $(1+nx)(1+x) ≤ (1+x)^n (1+x) $
2. $1 + x + nx + nx^2 ≤ (1+x)^{n+1}$
3. $1 + x + nx + nx^2 > 1 + x + nx$
4. So because of 3, $1 + (n+1)x$ must be less than or equal to $(1+x)^{n+1}$

I follow until number 3. I'm not sure why because that is true, it makes the inequality true using $(k+1)$, or $(n+1)$ in this case?

Was the problem done wrong?

Answer 1

An idea to complete the proof:

You assume $\;(1+x)^n\ge 1+nx\;$ , and you have to prove

$$\color{red}{(1+x)^{n+1}\ge1+(n+1)x}$$

Let us take the left side and develop it using the inductive hypotheses:

$$(1+x)^{n+1}=(1+x)^n(1+x)\stackrel{\text{Ind. Hyp.}}\ge(1+nx)(1+x)$$

So it is enough to prove

$$(1+nx)(1+x)\ge1+(n+1)x\iff\color{green}{1+x+nx}+nx^2\ge\color{green}{1+x+nx}$$

and observe the last inequality is immediate as $\;nx^2\ge 0\;$ .

Answer 2

Inductive step: $$ (1+x)^n \geq 1+nx $$ Proof step(?) $$ (1+x)^{n+1} =(1+x)^n (1+x) \geq (1+nx)(1+x) = 1+nx +nx^2+x \geq 1+nx +x $$ The first inequality is by the inductive step, the second is due to the positivity of $nx^2$.

Answer 3

Let me just write what you wrote with minor variations and extra explanations.

First the inequality is true when $n=1$ since then we have $1+x\ge1+x$. (You teacher says $n\ge2$ but it is also true for $n=1$ or even for $n=0$, I prefer to start with $n=1$ it is easier.)

So now using induction, assume it is true for some $n\ge1$ and prove it for $n+1$.

By induction hypothesis we have $(1 + x)^n \geq 1 + nx$. Use that $x\ge -1$, so $1+x\ge0$ and we may multiply both sides of the inequality by $1+x$, preserving the inequality sign (remember if we multiplied by something negative then we would have to reverse the inequality sign).
$(1 + x)^n(1+x) \geq (1 + nx)(1+x)$, so then
$(1 + x)^{n+1} \geq 1 + x + nx +nx^2 = 1 + (n+1)x +nx^2\ge 1 + (n+1)x$, where we dropped the positive term $nx^2$. Thus we have $(1 + x)^{n+1} \ge 1 + (n+1)x$ which is exactly what we needed to prove since this is the version of our inequality for $n+1$, which we proved assuming it was true for $n$.

Answer 4

First note that we can use the stronger relation $>$ than $\geq$ for the given inequality when we assume $x\neq 0$ (this inequality is actually known as Bernoulli's inequality). I will try to outline a clear proof below.

Claim: 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 $S(k+1)$, a subtlety regarding inequalities is address: If $a>b$ and $c>0$, then $ac>bc$; if $c<0$, then $a>b$ implies $ac<bc$. Here if the proof of $S(k+1)$: \begin{align} (1+x)^{k+1} &= (1+x)^k(1+x)\\ &> (1+kx)(1+x) \tag{since $x+1>0$}\\ &= 1+x+kx+kx^2\\ &= 1+(k+1)x+kx^2\\ &> 1+(k+1)x\tag{because $x\neq 0$} \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.