Showing that $1+a>0 \implies (1+a)^n \ge 1 + na$

Question

Establish the Bernoulli Inequality if $1+a>0$, then $(1+a)^n\ge 1+na$ for $n\ge 1.$

So far I have

$$(1+a)^k+1 \ge 1+(k+1)a,$$

but I don't know what to do from here to make the left and right side equal. Step by step explanation please!

already treated [ http://math.stackexchange.com/questions/46562/bernoulli-inequality ] — janmarqz, Jan 30 '14 at 03:15

score 2 · Answer 1 · answered Jan 30 '14 at 03:16

2

Suppose that $(1+a)^k \ge 1+ka$. Then $$(1+a)^{k+1} =(1+a)^k(1+a) \ge (1+ak)(1+a) = 1+a(k+1)+a^2k $$

Shazam!

answered Jan 30 '14 at 03:16

marty cohen

in your first step why is the left side 1+ka? because if you distribute a to (k+1) it is ka+a – Lil Jan 30 '14 at 03:17
ok and where did the 1+ka on the right come from?? – Lil Jan 30 '14 at 03:20
He is doing it by induction: if you suppose what you want for $k$, then you can show it is still true for $k+1$. On the line in the middle, the first inequality uses the assumption that what you want is true "at step $k$". – Clement C. Jan 30 '14 at 03:21
when you plug in k+1 on the right side wouldn't it be 1+a(K+1) which equals 1+ka+a?? – Lil Jan 30 '14 at 03:26
Which right side? For the proof by induction, you assume $$(1+a)^k \geq 1+ka$$ From there, you can only use this assumption; that's why you write $$(1+a)^{k+1}=(1+a)^{k}(1+a)$$ since this allows you to use you only tool (the assumption) on the first factor $(1+a)^{k}$. This leads to $$(1+a)^{k+1}\geq(1+ka)(1+a)$$ and now you can expand $(1+ka)(1+a)$ into $1+(k+1)a+a^2k$. Finally, since $a^2k \geq 0$, you can conclude by saying $$(1+a)^{k+1}\geq 1+(k+1)a+a^2k \geq 1+(k+1)a$$ – Clement C. Jan 30 '14 at 03:35

score 1 · Answer 2 · answered Jan 30 '14 at 03:18

1

The best way to tackle this problem is by induction on $n$. For instance, if $n = 1$, then

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

Suppose now $(1 + a)^n \geq 1 + an $. Then

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

The problem is now solved by induction.

answered Jan 30 '14 at 03:18

ILoveMath

how did you go from 1+an on the right side of the equation to (1+a)^n(1+a) wouldn't it be 1+a(n+1) which equals 1+an+a?? – Lil Jan 30 '14 at 03:23

score 0 · Answer 3 · answered Jan 30 '14 at 03:17

By induction on $n\geq 1$: fix any $a\geq 0$.

for $n=1$, it is true: $$(1+a)^1=1+a\geq 1+1\cdot a$$
assume the property holds up to some $n\geq 1$. Then, $$(1+a)^{n+1}=(1+a)(1+a)^{n}\operatorname*{\geq}_{\rm IH} (1+a)(1+na) = 1+(n+1)a + na^2 \geq 1+(n+1)a$$ so by induction, the property holds for any integer $n\geq 1$.

3 Answers3