Prove an inequality using induction

Question

I have to prove that...

$(1 +a)^n \ge 1+an$ for $ a > 0$ and $n \ge 1$

I've started with the following base case:

Let $a = 1$ and $n = 1$. Then $(1 +1)^2 \ge 1+(1)(1)→ 4\ge 2$, which is true.

However, I'm stumped after this point. I've tried to come up with a start to an inductive step, listed below; however, I don't believe it's correct.

Let $k ≥ 1$ and $b > 0$. Assume that $(1 +b)^k \ge 1+bk$.

Any help will be much appreciated.

Answer 1

Claim :

$$(1+a)^n\ge 1+an$$ for $a\ge -1$ , $n\ge 1$

Base Case : $n=1$

$$1+a\ge 1+a$$

obviously true

Assume $(1+a)^n\ge1+an$

Now, we have $$(1+a)^{n+1}=(1+a)^n\times(1+a)\ge(an+1)(a+1)=a^2n+an+a+1\ge a(n+1)+1$$ because of $a^2n\ge 0$, completing the proof.

Answer 2

Hint

Assume $a>-1$ and $(1+a)^n\ge 1+an.$ Then

$$(1+a)^{n+1}=(1+a)(1+a)^n\underbrace{\ge}_{\mbox{hypothesis}} (1+a)(1+na)=1+(n+1)a+na^2\ge 1+(n+1)a.$$

Answer 3

Try using the binomial expansion of (1+a)^n

Answer 4

$$(1+a)^n=\sum_{k=0}^n\binom{n}{k}1^{n-k}a^k=\binom{n}{0}1^na^0+\binom{n}{1}1^{n-1}a+\sum_{k=2}^n\binom{n}{k}1^{n-k}a^k\geq\binom{n}{0}1^na^0+\binom{n}{1}1^{n-1}a=1+na$$