Why do most inequalities require the condition for variables to be positive.

Question

I just struck with a doubt today

Why do most of the standard inequalities require the variables to be positive.

For example if we want to find minimum value of a certain expression say $a+b+c$ the very first thought that comes in our mind is the AM GM inequality but the question must satisfy a condition $\mathbf {a, b, c\ge 0}$.

So I want to ask why is that so.

Even in some very useful inequalities like the Muirhead inequality, Hölder's inequality, Minkowski's inequality, etc. we need the condition that the variables to be used must be non negative or positive.

While there are also some inequalities like Chebyshev's inequality and Rearrangement inequality, Cauchy Schwartz inequality which do not have restrictions of the variables or terms to be positive.

I want to know why such condition is needed to make the inequalities true. Is there a mathematical sense and a reason to do so? Does this have to do anything with geometry( I saw proofs of AM GM inequality using geometry and as the variables used were the lengths of some segments they were confined to be non negative)

If someone has an idea please share.

Answer 1

I would say this depends a lot on how you state things / what you interpret as standard form, and the proof you use. Had half a mind to vote to close as ambiguous, but will respond below to share my two cents in more detail.

One approach to look at this is how we state the inequality - arguably replacing variables with their squares allows the domain to be extended to all reals - e.g. the simplest AM-GM relation can either be stated as $a^2+b^2 \geqslant 2ab$ which follows from $(a-b)^2 \geqslant 0$, and of course holds for all reals $a, b$. However if we consider the standard form as $a+b \geqslant 2\sqrt{ab}$, then clearly we have specialised to non-negatives. Similarly we can write two forms for CS, etc. or in any inequality, square all variables which are non-negative to generalise to all reals. Clearly the form which has become popular matters.

Many inequalities in general stem from convexity - e.g. Jensen, which actually has no restriction on domain per se. However the function one uses in Jensen to prove a specific inequality e.g. if we specialise to $\log$ for AM-GM, the domain introduces the restriction. OTOH, inequalities which stem from considerations of order - such as Rearrangement, Abel etc. usually do not have these restrictions, unless specialised again to specific domains. An example would be Schur. So hard to say this is a generic reason - convexity or order in general poses no restrictions.

Some inequalities end up being applied in probability / measure contexts where the variables are expected to be non-negative, and the popular forms are thus specialised to non-negatives.

Further, there are "standard" inequalities which have other restrictions - e.g. Bernoulli, $(1+x)^r \geqslant 1+rx$ which has specific conditions on both $x, r$ and in fact reverses if the exponent $r \in (0, 1)$.

Not sure if any of that helps you with a conclusive answer, or even if one exists, but clearly too long for a comment.

Answer 2

Because AM-GM is not true for negative variables.

For example, $$\frac{x^3+y^3+z^3}{3}\geq xyz$$ is not true for $x=y=z=-1$

It's also Muirhead because $(3,0,0)\succ(1,1,1).$

AM-GM follows from Jensen for $\ln$, but domain of $\ln$ it's $(0,+\infty).$

Rearrangement, Chebyshov and C-S we can prove for all real variables.

Holder is not true for negative variables.

The proof of Holder follows from convexity of $f(x)=x^k$, where $k\geq1$.