Application of inequality in other math fields

Question

I'm learning inequalities for the first time, and except a paragraph by Paul Zeitz in his book Art and Craft of problem solving, none actually give much motivation of why should I care about inequalities.

The example given by Paul Zeitz was that to prove $b^2-b+1$ is never a perfect square for integer $b$. Well- that kinda motivates a tiny bit, but the inequality used is triviality obvious; I want much deeper.

What are some easy-to-state, moderately hard to solve number theoretic or combinatorics problem which requires applying a nontrivial inequality ?

Answer 1

The pigeonhole principle depends on an inequality and is nontrivial because it is so useful. Another is $x^2 \ge 0$ for $x$ real which can be used to prove some polynomials have no real roots.

Answer 2

(I). Many Q's are readily answered by the following: Let $J$ be a (bounded or unbounded) interval of $\mathbb R.$ Let $f:J\to \mathbb R$ with $f''(x)<0$ for all $x\in J.$ Then when $w_1,...,w_n$ are non-negative with $\sum_{i=1}^nw_i=1,$ and $x_1,...,x_n \in J$ then $$(*)\quad f\left(\sum_{i=1}^nw_ix_i\right)\geq \sum_{i=1}^nw_if(x_i)$$ with equality iff $x_i=x_j$ whenever $w_i\ne 0\ne w_j.$

Example: $J=(0,\infty)$ and $f(x)=\ln x.$ Then $\log (\sum_{i=1}^nw_ix_i)\geq \sum_{i=1}^nw_i\log x_i,\;$ which is equivalent to $\sum_{i=1}^n w_ix_i\geq \prod_{i=1}^nx_i^{w_i}.$

In particular when $w_i=1/n$ for each $n$ we have $(\sum_{i=1}^nx_i)/n\geq (\prod_{i=1}^nx_i)^n,$ which is the AGM inequality.

If, instead, we have $f''(x)>0$ for all $x\in J$ then the inequality in (*) is reversed. An example Q that was on this site this month (May 2017) was to minimize $\sum_{i=1}^nix_i^2\;$ given that $\sum_{i=1}^nix_i=1,$ for $x_1,...x_i\in \mathbb R.$ With $f(x)=x^2$ and $J=\mathbb R,$ let $w_i=i/((n^2+n)/2)$ for each $i$, and the answer is obtained immediately.

(II). On a different subject we have : Let $x\in \mathbb R.$ If for every $r>0$ there exist $a,b\in \mathbb Z$ such that $0<|x-\frac {a}{b}|<\frac {r}{|b|}$ then $x\not \in \mathbb Q.$

PROOF: If $x=c/d$ with $c,d\in \mathbb Z$ then for $a,b \in \mathbb Z$ we have $$0<|x-\frac {a}{b}|\iff 0<|cb-ad| \iff 1\leq |cb-ad| \iff$$ $$\iff \frac {1/|d|}{|b|}\leq |\frac {c}{d}-\frac {a}{b}|=|x-\frac {a}{b}|$$ so we cannot have $0<|x-\frac {a}{b}|<\frac {r}{|b|}$ unless $r\geq \frac {1}{|d|}.$

Note how the inequality $0<|cb-ad|$ strengthens to $1\leq |cb-ad|$ because $cb-ad\in \mathbb Z.$