One of the topics of my algebra subject is Proof by Induction. When it comes to proving equalities, whether they are a series, factorial or other types I had no problem doing so. But when it came to doing the same for inequalities I got stuck. I tried to watch videos on its proving process, read the answers I could find in this site, but there's something I'm not grasping as I'm not capable of finishing the proof I'm working on.

That's the following, given: $$ 2^n > n^2 + 4n + 5 \ for \ n > 6 \ where \ n \in \Bbb N $$

First, I compacted the expression to this: $ 2^n > (n + 2)^2 + 1 $

Then, I did the: $ \pmb{Base \ Case}: $ $$ P(7): 2^7 > (7 + 2)^2 + 1 = 128 > 82 $$

Which results to be True.

Then I established the following $\pmb{assumption} $ to be True for $ n = k $: $$ P(k): 2^k > (k+2)^2 + 1 \ where \ n > 6 $$

And went on to $\pmb{prove}$ for $ n = k + 1 $: $$ P(k + 1): 2^{k+1} > ((k+1)+2)^2 + 1 $$

So here started with the $ LHS = 2^{k+1} = 2(2^k)$

Then proceeded with the $ RHS = ((k+1)+2)^2 + 1 = (k+3)^2 + 1 $

But, as all the explanation I got about the Induction principle has been about its generic steps and using equalities, I don't really know how to proceed or what steps to follow to accomplish the proof nor what to do with the expressions I have.

So what I ask for is an explanation of how should I proceed when I encounter an equality like the one I'm stuck with, because If I only get the answer now, later on, I'll be equally stuck.

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    $\begingroup$ To show the inequality for $P(k+1)$, you want to somehow use $P(k)$. A natural thing to do could be to start with $P(k)$ (which we assume is true) and multiply both sides by $2$ (since this makes the LHS become $2^{k+1}$, so it resembles the $P(k+1)$). Then try and simplify the RHS down to what is desired. $\endgroup$ – Minus One-Twelfth Mar 6 '19 at 11:30
  • $\begingroup$ Thanks, @MinusOne-Twelfth, this helped. That's a great piece of advice. That's the kind of thing I'm looking for. The set of things that would be considered "natural" on resolving this kind of problems. $\endgroup$ – Power_of_zero Mar 6 '19 at 13:19

We have $$2^n>n^2+4n+5$$ and we must prove that $$2^{n+1}>(n+1)^2+4(n+1)+5$$ multiplying our first inequality by $2$ we get $$2^{n+1}>2n^2+8n+10$$ so it remains to prove that $$2n^2+8n+10>(n+1)^2+4(n+1)+5$$ This is true, since we get $$n(n+2)>0$$

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    $\begingroup$ Thanks! Combining your answer with the comment in the question, I see how you got where you got. This helped me in resolving the problem, but, as I already did in a comment above, I would kindly ask for some set of thoughts or ideas to keep in mind when encountering this kind of problems. $\endgroup$ – Power_of_zero Mar 6 '19 at 13:21

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