How to proceed with induction step of induction proof

Question

$$ P(n): (1+a)^n \geq 1 + a(n - 1) \\ a \geq 0 $$

I already proved its base case with n = 1, which is pretty simple. However, I am confused on how to proceed with my induction step of

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

Any tips on what to do next? Thanks

Answer 1

You must assume $a \ge 0$ for the case $n=1$ to be true.

Note we must prove $(1+a(n-1))(1+a) \ge 1+a(n)$. Let $n=t+1$.

This implies that we must prove that $a^2t + a(t+1) +1 \ge 1+a(t+1)$, which follows from $a \ge 0$, $t \ge 0$.

Answer 2

Assume $P(n)$ is true as per the induction hypothesis.

Hence $(1+a)^n \ge 1+a(n-1)$

Now for $P(n+1)$,

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

since $a^2(n-1)$ is always positive for $n > 1$.

Hence $P(n+1)$ is true.