# Unfamiliar inequality rules in proofs. (learning proofs)

Without going into too much detail, I don't think we actually went over inequalities in much depth in any crappy high school or college math class I've ever taken. Now that I am studying proofs, I notice a constant use of inequality tricks without justification (so I am assuming its supposed to be either obvious or known to be true.)

Over time I sort of intuitively or habitually picked up on some patterns, but I want something more concrete.

Looking online for inequalities only gives me few very basic rules. Where do I find a more detailed source for knowing and understanding more elaborate algebraic inequality arguments.

Yeah to clarify I am talking about rational function, absolute value inequalities etc... I am sure it is basic stuff for most of you.

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As a side note, I also see this occur a fair amount in measure theory, which I am currently studying. –  Jebruho Jan 29 '13 at 7:19
You can try Googling 'famous inequalities' or other. This comes up, for instance: 2000clicks.com/MathHelp/IneqMethod.aspx –  ido Jan 29 '13 at 7:22
You should specify what kind of inequalities you are referring to. There's a big difference between rational function inequalities $\frac {x+1}{x+2} \leq 3$, working with absolute values $|x+1| > |2x-4|$ and trying to apply AM-GM, Cauchy Schwarz, Holders, etc. –  Calvin Lin Jan 29 '13 at 7:22
@CalvinLin I am talking about the former. (rational, absolute) –  Eugene Jan 29 '13 at 7:54
May I recommend that you keep going? The more you study, the more you will learn the technique. I don't think it is very productive to pause and learn the technique alone. –  Giuseppe Negro Jan 29 '13 at 19:35

This is a great book especially suited for self study. The Cauchy-Schwarz Master Class...

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Many of the simpler inequalities are of the form $f(x)<g(x)$ where $f,g$ are continuous functions (or the $<$ replaced by $\le,\ >,\ \ge$). In such a case one can work with equations instead: Let $A$ be the set of reals for which $f(x)=g(x)$ (where both sides are defined), and let $B$ be the set of reals for which one or both of $f(x),g(x)$ is undefined. Now remove from the real line all pointsw in sets $A,B$, and then the line divides into a collection of intervals. Then you only need to check whether $f(x)<g(x)$ is true or not on each interval (which only needs a check at any convenient point in such an interval). Once this is known, you can include the set $A$ in the answer provided the inequality was inclusive ($\le,\ \ge$).

I've found this method works quite well, since it seems easier to solve equations or determine undefined points than to solve inequalities by looking at various cases, etc.

Of course this method may not be considered proper in cases where one is trying to prove inequalities on the way toward showing $f$ or $g$ is continuous, since this method assumes they are known to be continuous already.

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Great question.

Claim. As long as we're dealing with monotonic functions, we can think of an order relation like $\leq$ as just being a directed version of equality.

For example, suppose $x$ and $y$ are positive real numbers that and $z$ is a real number. Furthermore, assume the following. $$z-x \leq xy,\quad x+(xy)^2 \leq z-1.$$

What can we deduce from this? Well, if the above were both just equalities, then we could deduce $$x+(z-x)^2 = z-1.$$

The important observation is that even though they aren't equalities, we can nonetheless deduce

$$x+(z-x)^2 \leq z-1.$$

Proof. Each line in the following list implies the next.

1. $z-x \leq xy$
2. $(z-x)^2 \leq (xy)^2$
3. $x+(z-x)^2 \leq x+(xy)^2$

But the RHS is less than or equal to $z-1$, so

$$x+(z-x)^2 \leq z-1.$$

Cool, right?

If I have the time, I'll explain why this happens in more detail.

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