Nested hypotheses in proof by induction

Question

Background

I have to prove that, given positive numbers $x_k$

$$ \prod_{k=1}^n x_k = 1 \Longrightarrow \sum_{k=1}^nx_k \geq n $$

where $\Longrightarrow$ means that the first condition implies the second. I already know this can be proved as a trivial sub case of the AM-GM inequality.

Here is my question:

I know this can also be proved by induction, however I don't understand how one can use the inductive hypothesis if this has in turn another hypothesis nested in it. In other words:

If I call P(n) the proposition $ (\prod_{k=1}^n x_k = 1 \Longrightarrow \sum_{k=1}^nx_k \geq n )$ ("the first implies the second") than the proof by induction would work like this

1. Verify $P(0)$
2. Show that $P(n) \Longrightarrow P(n+1)$

where $P(n)$ is the proposition defined above.

How would one prove the inequality this way? I find very confusing the idea of showing that "the first implication implies the second implication".

Answer 1

Implications implying implications should, as a general idea, not really be any more difficult as proving any other implication. For example, the implication$ P \to Q$ implies the implication $(P \land R) \to Q$. How do you prove something like that? Well, you first assume the implying implication to be true:

1. Assume $P \to Q$

Next, assume the antecedent of the (allegedly) implied implication:

2. Assume $P \land R$

Now try to prove consequent of the implied implication, which is $Q$. In this case that is easy:

3. $P$ (follows immediately from 2)

4. $Q$ (follows from 1 and 3)

Ok, so now we can wrap up the inside consitional proof:

5. $( P \land R) \to Q$ (From 2 through 4)

And then wrap up the outside conditional proof:

6. $P \to Q \implies (P \land R) \to Q$ (From 1 through 5)

Ok, so use this same proof structure for your proof!