Help with an inductive proof of $(1+a)^n\ge 1+na$

Question

Let $a$ be a fixed real number such that $a > -1$. Prove $(1+a)^n \geq 1+an$ for any $n\in\mathbb{N}$.

Base case. $n = 1$.

$(1+a)^1 = 1+a(1)$

$\implies 1+a = 1+a$.

$1+a = 1+a(1)$.

so we can assume $P(n)$ is true when $n = 1$.

Want to show $P(n+1)$ holds.

$(1+a)^n = 1+an$

$(1+a)^n(1+a) = (1+an)(1+a)$

$(1+a)^{n+1} = 1+a+an+a^2n$

Notice $a^2n \geq 0$ since for $a > -1, a^2 > 0$.

$(1+a)^{n+1} \geq 1+a+an$

$(1+a)^{n+1} \geq 1+a(n+1)$.

so $P(n+1)$ holds.

Can someone scrutinize my base case? If I use $n=1$, how can I show that it's greater than or equal to, since the base case is just equal?

@symplectomorphic is correct. All you need to prove is that your statement is true for the base case, you don't need to (and can't) prove that it is both greater AND equal. — rb612, Jun 03 '16 at 04:50
@rb612: I'd be interested to see anyone prove that one number is both "greater AND equal" to another. — symplectomorphic, Jun 03 '16 at 04:51
@symplectomorphic LOL, that was what I meant. You can't do so :) — rb612, Jun 03 '16 at 04:52

score 3 · Accepted Answer · answered Jun 03 '16 at 05:27

If you know $A=B$, you know that $A \le B$ and $A \ge B$.
In fact, $(A = B) \iff (A \le B) \land (A \ge B)$ Therefore, your base case becomes:
$n = 1$.

$(1+a)^1 = 1+a(1) = 1+a$
$\implies(1+a)^1 \ge 1+a(1)$
$\therefore P(1)$ holds

For the rest of the proof:

$(1+a)^n \ge 1+an$
$\implies (1+a)^n(1+a)\ge (1+an)(1+a)$
$= 1+an + a + a^2n$
$= 1+a(n+1)+a^2n$

$a^2 \ge 0, \forall a \in \Bbb{R}$
$\therefore 1+a(n+1)+a^2n \ge 1+a(n+1), \forall a\in \Bbb{R}, \forall n \ge 0$
$\therefore P(1) \land (P(n) \implies P(n+1))$

$\therefore P(n), \forall n \in \Bbb{N}$

Oh I get it so that's where the thing about that little $a^2k$ part being positive comes into play. — J00S, Jun 03 '16 at 05:31

score 2 · Answer 2 · 2016-06-03T11:34:12.770

The case $n=1$ has already been addressed in the comments.

A couple of further issues in your work:

It is not true that if $a>-1$, the $a^2>0$. In fact, $0>-1$. However, $a^2n\ge0$ holds regardless of any hypothesis stronger than, say, $n\in\Bbb N$ and $a\in \Bbb R$ (and "$a^2n\ge0$" is sufficient for the purpose of the proof). The point where you use $a>-1$ is just in the the implication $$(1+a)^n\ge 1+na\implies (1+a)^{n+1}\ge(1+a)(1+na)$$
... which leads us to your second imprecision: the inductive step that you must prove is $$\forall n\ge1,\ \left[(1+a)^n\ge1+na\longrightarrow (1+a)^{n+1}\ge 1+(n+1)a\right]$$ while, formally, you proved the weaker predicate $$\forall n\ge1,\ \left[(1+a)^n=1+na\longrightarrow (1+a)^{n+1}\ge 1+(n+1)a\right]$$ The algebra is the same as you did, though, just with the "$\ge$" sign instead of "$=$".

Help with an inductive proof of $(1+a)^n\ge 1+na$

2 Answers2