In the induction proof for $(1+p)^n \geq 1 + np$, a term is dropped and I don't understand why. In What is Mathematics, pg. 15, a proof of 
$(1+p)^n \geq 1 + np$, for $p>-1$ and positive integer $n$
goes as follows:


*

*Substitute $r$ for $n$, then multiply both sides by $1+p$, obtaining:
$(1+p)^{r+1}\geq 1+rp+p+rp^2$

*"Dropping the positive term $rp^2$ only strengthens this inequality, so that
$(1+p)^{r+1}\geq 1+rp+p$, which shows that the inequality will hold for $r+1$."
I don't understand why the $rp^2$ term can be dropped -- if we're trying to prove that the inequality holds, and dropping $rp^2$ strengthens the inequality, then why are we allowed to drop it?
Thanks!
 A: We know that $$(1+p)^{r+1}\ge 1+rp+p+rp^2\;.\tag{1}$$
We know that $rp^2\ge0$, which means that $$1+rp+p+rp^2\ge1+rp+p\;.\tag{2}$$
Put $(1)$ and $(2)$ together:
$$(1+p)^{r+1}\ge 1+rp+p+rp^2\ge1+rp+p\;.\tag{3}$$
And this is what we wanted to show: $(1+p)^{r+1}\ge1+(r+1)p$. The point is really just that if $(1)$ is true, then $(2)$ is certainly true, and since $(2)$ is what we want, we can ignore the fact that the stronger statement $(1)$ is also true.
A: One point I am missing in both the question and the other answers is the proper context: this is a proof by induction. That means that we assume the inequality for $n$, which is then renamed $r$ so that we can now focus on proving the same result for $r+1$ (which we can then call our new $n$). Therefore, before and after step 1. we are dealing with inequalities that are known to hold, and we can combine them with other inequalities like $rp^2\geq0$ in the usual fashion, as explained in other answers.
However if this had been a non-induction proof, then $(1+p)^n \geq 1 + np$ would just be our goal, and dropping positive terms from the right hand side would indeed be illegal, as this weakens the goal (we would end up proving less than we claimed). Incidentally, the multiplication by the positive factor $(1+p)$ in step 1. would still be justified (as $a\geq b$ if and only if $ra\geq rb$ when $r>0$), but for the opposite reason: in the actual induction proof we are saying "we assume $(1+p)^r \geq 1 + rp$ and this implies $(1+p)^{r+1} \geq 1 + rp+p+rp^2$" while in a non-induction proof we could validly argue "we need to prove $(1+p)^r \geq 1 + rp$, and this will follow from $(1+p)^{r+1} \geq 1 + rp+p+rp^2$" (but then we would get stuck, and in particularly dropping $rp^2$ would be disallowed).
By the way, the book is really wrong in saying that dropping $rp^2$ strengthens the inequality, it really weakens it, as I said above. But what the author probably wanted to say is that dropping the term only makes the inequality more (easily) true, with a greater distance between the two members.
A: We can drop it, because the inequality still holds true if we drop it. We are not proving the inequality here, inequality is already assumed true. So we just derive a weaker inequality from a stronger one, because the weaker version is the one that is useful for our purpose.
Of course if we were trying to prove the original inequality, then we could not drop terms from it, prove the weak version and then say that we proved the original inequality.
A: In $1.$ we have shown that $$(1+p)^{r+1}\geq 1+rp+p+rp^2$$
But we also know that $r > 1$ (because we're doing an induction proof from $1$ upwards); and obviously $p^2 \ge 0$ (because $p$ is real); so we know that $rp^2 \ge 0$. Therefore
$$1+rp+p+rp^2 \ge 1+rp+p$$
So putting these two together gives
$$(1+p)^{r+1}\geq 1+rp+p$$
as required.
In short, if we know that $a \ge b + c$, and we know $c$ is non-negative, we can immediately conclude that $a \ge b$.
A: if $a \geq b$ and $b \geq c$, then  $a \geq c $. Since $rp^2$ is positive then it is clear that $x+ rp^2 > x $   
