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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?

John
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4 Answers4

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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 $\;nx^2\ge 0\;$ .

Timbuc
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  • I'm just confused by this part: ≥Ind. Hyp.(1+nx)(1+x). SHouldn't that side be 1+(n+1)x? – John Jan 23 '15 at 21:57
  • And of course your must also check the base case $n=2$, which no one mentioned here. – user26486 Jan 23 '15 at 21:57
  • Ya I've done all the steps up to this point. The induction proof is just whats confusing me. – John Jan 23 '15 at 21:58
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    @John You want to prove that $(1+x)^n(1+x)\ge 1+(n+1)x$, but you don't know this yet. What you do know though is that $(1+x)^n\ge 1+nx$ and thus $(1+x)^n(1+x)\ge (1+nx)(1+x)$. Furthermore, $(1+nx)(1+x)\ge 1+(n+1)x$. Thus you have proved that $(1+x)^{n+1}=(1+x)^n(1+x)\ge (1+nx)(1+x)\ge 1+(n+1)x$ and thus $(1+x)^{n+1}\ge 1+(n+1)x$, which is what we wanted to prove. – user26486 Jan 23 '15 at 21:58
  • @John: you want to reach the inequality in red. So we take that left side and develop it. At the end we get that it certainly is greater than or equal to the right side in red, as wanted. – Timbuc Jan 23 '15 at 21:59
  • Oh so because the left side ends up being (1+x)^n+1, we have proved it? – John Jan 23 '15 at 22:02
  • @John NO, but almost: the left side ends up being $;\ge;$ than $;1+(n+1)x;$ , and this is what we wanted. – Timbuc Jan 23 '15 at 22:04
  • Ohhh but since the RHS is (1+nx)(1+x) > 1+(n+1)x, its been proven. – John Jan 23 '15 at 22:07
  • @John Yup, that is. – Timbuc Jan 23 '15 at 22:12
  • Okk thanks guys. My teacher is terrible at explaining this so this has been a huge help. – John Jan 23 '15 at 22:15
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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 $nx^2$.

Alex
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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 $n$.

Mirko
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  • Okk this all starting to make since. My only questions is did we drop the nx2 because we already new the LHS was bigger, and now we have the exact thing we are trying to prove? – John Jan 23 '15 at 22:10
  • we dropped $nx^2$ for two reasons: 1. since $nx^2\ge0$ (since both $n\ge0$ and $x^2\ge0$) so we are allowed to drop them, extending the inequality in the "correct" direction, and 2. since once we do that, we have exactly what we needed to finish the proof. – Mirko Jan 23 '15 at 22:13
  • Ok I think I finally understand this. Thanks for the help. – John Jan 23 '15 at 22:16
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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.

Daniel W. Farlow
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