Techniques for proving inequalities

Quick description

This page contains links to articles that discuss techniques for proving inequalities.

The articles

Compressions Quick description ( Suppose that a parameter $\phi$ (such as volume, for instance) is associated with a certain kind of mathematical object, and one would like to find the object $X$ for which $\phi(X)$ is maximized. A useful general idea is to find a compression: that is, an operation $\kappa$ that takes an object $X$ and produces a new object $\kappa(X)$ with the property that $\phi(\kappa(X))$ is at least as big as $\phi(X)$ , with equality if and only if $\kappa(X)=X$ . Then we know that any $X$ for which $\phi(X)$ is maximized must satisfy $\kappa(X)=X$ , which may well give us enough information about $X$ to allow us to solve the problem. )

Proving inequalities using convexity Quick description ( Jensen's inequality states that if $f$ is a convex function and $a_1,\dots,a_n$ are positive real numbers that add up to 1, then $f(a_1x_1+...+a_nx_n)\leq a_1f(x_1)+...+a_nf(x_n)$ . Many well-known inequalities follow from this one. More generally, convexity can be a powerful tool for proving inequalities. )

Double counting. Quick description ( If you have two ways of calculating the same quantity, then you may well end up with two different expressions. If these two expressions were not obviously equal in advance, then your calculations provide a proof that they are. Moreover, this proof is often elegant and conceptual. Double counting can also be used to prove inequalities, via a simple result about bipartite graphs. )

Sums of squares Quick description ( If $A$ and $B$ are real numbers and you want to prove that $A$ is at most as big as $B$ , then often a good way to do it is to express $B-A$ as a sum of squares of real numbers. )

The tensor power trick Quick description ( If you want to prove that $X\leq Y$ , where $X$ and $Y$ are some non-negative quantities, but you can only see how to prove a quasi-inequality that says that $X\leq CY$ for some constant $C\geq 1$ , then try to replace all objects involved in the problem by "tensor powers" of themselves and apply the quasi-inequality to those powers. If all goes well, one can show that $X^M\leq CY^M$ for all positive integers $M$ , with a constant $C$ which is independent of $M$ , which implies that $X\leq Y$ since one can take $M$ th roots and then let $M$ tend to infinity. )

Bounding probabilities by expectations Quick description ( If $X$ is a random variable and you would like a good upper bound for the probability that $X\geq t$ , then a good way of doing so is sometimes to choose a non-negative increasing function $F$ and use the fact that $\mathbb{P}[F(X)\geq F(t)]$ is at most $\mathbb{E}F(X)/F(t).$ This article is a discussion of the method. )