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:


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*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 see this post is old, but I answer it in case another one has the same problem.
If you have to prove $$1 + 3 + 5 + \cdots + (2n-1) = n^2\tag{*}$$  it is enough to find the LHS of the previous equation in the LHS of $$1 + 3 + 5 + \cdots + (2n-1) + (2(n+1)-1)= (n+1)^2\tag{**}$$The next step is assuming the first equation is true and replace the RHS of (*) in LHS of (**) to obtain $$n^2+(2(n+1)-1)=n^2+2n+1$$ Algebraically, it simplifies as $$n^2+2n+1=n^2+2n+1$$ which proves the first statement.
If you have to prove an inequality holds, the trick is to find what you have on each side of (n) assumption on each side of (n+1)  assumption.
In the induction step of your example, you have $$1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{k}+\frac{1}{k+1}\le\frac{k+1}{2}+1\tag{1}$$
Which can be organized as $$(\frac{1}{k+1})+(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{k})\le(\frac{k}{2}+1)+\frac{1}{2}\tag{2}$$
I did not add anything, I just reorganized each side of the inequality to show that the statement $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{k} \leq \frac{k}{2} + 1$ is contained in (1).
Now, if I assume that $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \leq \frac{n}{2} + 1$ is true, then (2) must be true since $\frac{1}{k+2}\leq\frac{1}{2}$ for every positive integer.
