Is this a valid inductive step to prove Bernoulli's inequality inductively?

Question

I am wondering if the following is a valid inductive step for an inductive proof of Bernoulli's inequality, $ \forall n \in \mathbb{N},(1+x)^n \geq 1+nx, $ with $ x \geq -1$. Here it is:

1. We endeavor to show that $(1+x)^{k+1} \geq 1+(k+1)x$ using the fact that $(1+x)^k \geq 1+kx$, our inductive hypothesis.
2. The former inequality is $(1+x)(1+x)^k \geq 1+kx+x$
3. We can reason that 2. holds based on the inductive hypothesis. The inductive hypothesis can be transformed into 2. by multiplying its left-hand-side by $(1+x)$, and adding $x$ to its right-hand-side. If $x$ is negative, clearly 2. holds because we are subtracting a number, $x$, from a smaller/equal quantity ($1+kx$), and multiplying a bigger/equal quantity ($(1+x)^{k+1}$) by a positive quantity on the right ($1+x$ is always positive). If $x$ is positive, we can see that we have increased the right of the inductive hypothesis by a greater order of magnitude than we have the left by transforming it thus. Take $a\ge b$ where $a\gt0$, and transform it the same way we can transform the inductive hypothesis to get 2. like so: $(x+1)a\ge b+x$, which is $a+ax\ge b+x$. We can see this inequality with $a$ and $b$ holds because each of the terms on the left are greater than or equal to the terms on the right. We can apply this fact to the inductive hypothesis transforming into 2., to see that the inequality in the inductive hypothesis has been preserved.

Is this mathematically and logically valid?

Answer 1

Note: Changed after clarification

The inductive hypothesis can be transformed into 2. by multiplying its left-hand-side by (1+x), and adding x to its right-hand-side. If x is negative, clearly 2. holds because we are subtracting a number, x, from a smaller/equal quantity (1+kx), and multiplying a bigger/equal quantity ((1+x)k+1) by a positive quantity on the right (1+x is always positive).

That isn't a correct conclusion. If $x$ is negative, then you are right that adding $x$ to the right of the induction hypothesis makes it smaller. But so does multiplying the left by $(1+x)$, which is less than $1$. So both operations makes their sides smaller, and you have given now reason to assume which side gets more affected.

Multiplying both sides of a valid inequality by a non-negative number yields a new inequality that is also valid. Try multiplying both sides of the induction hypothis with $(1+x)$. That gives you the left side of what you want to proof, and with a bit of algebra on the right side you should be able handle the right side.