Quadratic Inequalities on [-1,1]

By substituting $x=1,x=-1,x=0,\,$ $|a+b+c|\le p,\,$ $|a-b+c|\le p,$ $|c|\le p.$

Now,

$\begin{align} 2|a+c|&=|(a+b+c)+(a-b+c)|\\ &\le |a+b+c|+|a-b+c|\le 2p, \end{align}$

so that $|a+c|\le p.$ Further,

$\begin{align} 2|b|&=|(b+a+c)+(b-a-c)|\\ &\le |a+b+c|+|a-b+c|\le 2p, \end{align}$

so that $|b|\le p.$ Finally, for $|x|\le 1,$ if $c$ and $a$ are of the same sign then

$|cx^2+bx+a|\le |c+a|+|b|\le 2p.$

If they are of different signs then

$|cx^2+bx+a|\le ||c|-|a||+|b|\le 2p,$

provided $|c|\ge |a|,$ because then $||c|-|a||\le |c|\le p.$

The same argument holds when $|c|\le |a|\le 2|c|.$ However, these cases do not exhaust all the possibilities, meaning that the proof is

... to be continued ...

Let $f(x)=ax^2+bx+c$ and $g(x)-cx^2+bx+a.$ It is given that

$\begin{align}|f(1)|&=|a+b+c|\le p,\\ |f(-1)|&=|a-b+c|\le p,\\ |f(0)|&=|c|\le p. \end{align}$

Then using the triangle inequality for the absolute value,

$\displaystyle \begin{align} |g(x)|&=|cx^2+bx+a|\\ &=\left|c(x^2-1)+(a+b+c)\cdot\frac{x+1}{2}+(a-b+c)\cdot\frac{1-x}{2}\right|\\ &\le |c||x^2-1|+|a+b+c|\cdot\left|\frac{x+1}{2}\right|+|a-b+c|\cdot\left|\frac{1-x}{2}\right|\\ &\le p\left(|x^2-1|+\frac{|x+1|}{2}+\frac{|1-x|}{2}\right)\\ &= p\left(1-x^2+\frac{x+1}{2}+\frac{1-x}{2}\right)\\ &=p(2-x^2)\le 2p, \end{align}$

for $x\in [-1,1].$

With $x$ taking values in $\{-1,0,1\}$, we have

$\left\{ \begin{array}{c} (a+b+c)^2\leq p^2 \\ c^2\leq p^2 \\ (a-b+c)^2\leq p^2 \\ \end{array} \right.$

and need to prove

$\left\{ \begin{array}{c} (a+b+c)^2\leq 4 p^2 \\ a^2\leq 4 p^2 \\ (a-b+c)^2\leq 4 p^2 \\ \end{array} \right.$

which simplifies to need to prove $a^2\leq 4 p^2$.

The space of solutions becomes:

$\begin{align} &b=-p\land -p\leq c\leq p\land a=-b-c-p \, \text{ (satisfied)}\\ &-p\lt b\leq 0\land -p\leq c\leq p\land -b-c-p\leq a\leq b-c+p \, \text{ (satisfied)}\\ &0\lt b\lt p\land -p\leq c\leq p\land b-c-p\leq a\leq -b-c+p)\, \text{ (satisfied)}\\ &b=p\land -p\leq c\leq p\land a=b-c-p)\, \text{ (satisfied)} \end{align}$

Dorin Marghidanu has kindly posted the above problem at the CutTheKnotMath facebook page and later commented with a solution of his (Solution 2). He also remarked that the problem with $p=1$ was published as #2153 in the Crux Mathematicorum (1996, p 217). Solution 3 is by N. N. Taleb.