How to use mathematical induction with inequalities? I've been using mathematical induction to prove propositions like this:
$$1 + 3 + 5 + \cdots + (2n-1) = n^2$$
Which is an equality. I am, however, unable to solve inequalities. For instance, this one:
$$ 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \leq \frac{n}{2} + 1 $$
Every time my books solves one, it seems to use a different approach, making it hard to analyze. I wonder if there is a more standard procedure for working with mathematical induction (inequalities).
There are a lot of questions related to solving this kind of problem. Like these:


*

*How to prove $a^n < n!$ for all $n$ sufficiently large, and $n! \leq n^n$ for all $n$, by induction? - in this one, the asker was just given hints (it was homework)

*How to prove $n < n!$ if $n > 2$ by induction? Ilya gave an answer, but there was little explanation (and I'd like some more details on the procedure)

*how: mathematical induction prove inequation Also little explanation. Solving it with one line is great, but I'd prefer large blocks of text instead.


Can you give me a more in depth explanation of the whole procedure?

I'm not sure what you expect exactly, but here is how I would do the inequality you mention. 
We start with the base step (as it is usually called); the important point is that induction is a process where you show that if some property holds for a number, it holds for the next. First step is to prove it holds for the first number. So, in this case, $n=1$ and the inequality reads
$$
1<\frac12+1,
$$
which obviously holds. 
Now we assume the inductive hypothesis, in this case that 
$$
1+\frac12+\cdots+\frac1n<\frac{n}2+1,
$$
and we try to use this information to prove it for $n+1$. Then we have
$$
1+\frac12+\cdots+\frac1n+\frac1{n+1}=\left(1+\frac12+\cdots+\frac1n\right)+\frac1{n+1}.
$$
I inserted the brackets to show that we have the sum we know about, through the inductive hypothesis: so
$$
1+\frac12+\cdots+\frac1n+\frac1{n+1}<\frac{n}2+1+\frac1{n+1}.
$$
Now comes the nontrivial part (though not hard in this case), where we need to somehow get $(n+1)/2+1$. Note that this is equal to $n/2+1$ (which we already have) plus $1/2$. And this suggests the proof: as $n\geq1$, $1/(n+1)\leq1/2$. So 
$$
1+\frac12+\cdots+\frac1n+\frac1{n+1}<\frac{n}2+1+\frac1{n+1}\leq\frac{n}2+1+\frac12=\frac{n+1}2+1.
$$
So, assuming the inequality holds for $n$, we have shown it holds for $n+1$. So, by induction, the inequality holds for all $n$. 
