Proof by induction of Bernoulli's inequality: $(1+x)^n \geq 1+nx$

Question

So all answers I've seen have to do with multiplying both sides by $(k+1)$ and continuing from there. The way I thought if it is different so I'd like to ask if it correct and if not why.

Note: I did look around the site a bit but I didn't find satisfying answers, sorry if it's a common question and I just missed answers.

My answer: After confirming its true for some $n$ values and assuming $(1+x)^k \geq 1 + kx$, I went like this:

I tried checking whether $(1+x)^{(k+1)} \geq 1 + (1+k)x$ is true

We have: $(1+x)(1+x)^k \geq 1 + x + kx$

Divide by $1+x$ as $x > -1: (1+x)^k \geq 1 + \dfrac{kx}{(1+x)}$

I add and subtract $kx$ and I have: $(1+x)^k \geq 1 + kx + \dfrac{kx}{(1+x)} - (1+x)\dfrac{kx}{(1+x)}$.

Now as we know $(1+x)^k \geq 1 + kx$ is true I just have to prove $\dfrac{((1+x)kx - kx)}{(1+x)}$ is greater than 0.

So: $\dfrac{(kx + kx^2 - kx)}{(1+x)} = \dfrac{kx^2}{(1+x)}$ which is greater than 0.

In other words $(1+x)^{(k+1)} \geq 1 + (1+k)x$ is true.

Notes: Sorry if I didn't explain something well enough, English isn't my mother tongue so explaining math in it is harder than simply talking.

Also, I understand the other method, I was just wondering if this is wrong.

Answer 1

We need $x>-1$ in the given.

Hence, by the assumption of the induction we obtain: $$(1+x)^{n+1}\geq(1+x)(1+nx)=1+(n+1)x+nx^2\geq1+(n+1)x$$

Answer 2

we use $$(1+x)^n\geq 1+nx$$ multiplying by $$(1+x)>0$$ we get $$(1+x)^{n+1}\geq (1+nx)(1+x)=1+x(n+1)+nx^2\geq 1+x(n+1)$$