Inequality on Root of Polynomials

Question

Suppose that $|a_i|<1$ for $i\in \{1,2,3,4\}$, and consider the polynomials

$$ x^2 - (a_1+a_2+a_3)x + a_1 a_2 + a_1 a_3 + a_2 a_3 - 1,$$

and

$$ x^2 - (a_1+a_2+a_3+a_4)x + a_1 a_2 + a_1 a_3 + a_1 a_4 + a_2 a_3+ a_2 a_4+ a_3 a_4 - a_1 a_2 a_3 a_4 - 1.$$

I want to show that, in both cases, any real root $x^*$ of these polynomials must satisfy $|x^*|<2$.

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Though it may look like it, this is not a homework question. It is something that appeared in my own research. It looks simple, but I've tried the usual avenues without luck. Any help would be much appreciated.

Answer 1

Let $f,g$ be given by \begin{align*} f(x)&=x^2-(a_1+a_2+a_3)x + a_1a_2+a_1a_3+a_2a_3 -1\\[4pt] g(x)&=x^2-(a_1+a_2+a_3+a_4)x+a_1a_2+a_1a_3+a_1a_4+a_2a_3+a_2a_4+ a_3 a_4 - a_1a_2a_3a_4 - 1\\[4pt] \end{align*} where $|a_i|<1$ for $i\in \{1,2,3,4\}$.

The goal is to show that any real root $r$ of either $f$ or $g$ satisfies $|r| < 2$.

First consider $f$ . . .

As noted in David's answer, the vertex of the parabola $y=f(x)$ has $x$-coordinate $$\frac{a_1+a_2+a_3}{2}$$ which lies in the interval $\left(-\frac{3}{2},\frac{3}{2}\right)$.

Hence to show that any real root $r$ of $f$ satisfies $|r| < 2$, it suffices to show that $f(-2) > 0$ and $f(2) > 0$.

For $f(-2)$, we have the identity \begin{align*} f(-2)&=3+2(a_1+a_2+a_3)+a_1a_2+a_2a_3+a_3a_1\\[4pt] &=(1+a_1)(1+a_2)+(1+a_2)(1+a_3)+(1+a_3)(1+a_1)\\[4pt] \end{align*} hence $f(-2) > 0$.

Similarly, for $f(2)$, we have the identity \begin{align*} f(2)&=3-2(a_1+a_2+a_3)+a_1a_2+a_2a_3+a_3a_1\\[4pt] &=(1-a_1)(1-a_2)+(1-a_2)(1-a_3)+(1-a_3)(1-a_1)\\[4pt] \end{align*} hence $f(2) > 0$.

This completes the analysis for $f$.

Next consider $g$ . . .

Note that the vertex of the parabola $y=g(x)$ has $x$-coordinate $$\frac{a_1+a_2+a_3+a_4}{2}$$ which lies in the interval $(-2,2)$.

Hence to show that any real root $r$ of $g$ satisfies $|r| < 2$, it suffices to show that $g(-2) > 0$ and $g(2) > 0$.

For $g(-2)$, we have the identity $$ g(-2) = 3 + 2(a_1+a_2+a_3+a_4) + a_1a_2+a_1a_3+a_1a_4+a_2a_3+a_2a_4+a_3a_4 - a_1a_2a_3a_4 $$ which is our target for the $\text{LHS}$ of the chain below . . . \begin{align*} &1-a_1a_2 > 0\\[4pt] \implies\;&(1-a_1a_2)a_3 > (1-a_1a_2)(-1)\\[4pt] \implies\;&a_3-a_1a_2a_3 > -1+a_1a_2\\[4pt] \implies\;&(2+a_1+a_2)+(a_3-a_1a_2a_3) > (2+a_1+a_2)+(-1+a_1a_2)\\[4pt] \implies\;&2+a_1+a_2+a_3-a_1a_2a_3 > 1+a_1+a_2+a_1a_2\\[4pt] \implies\;&2+a_1+a_2+a_3-a_1a_2a_3 > (1+a_1)(1+a_2)\\[4pt] \implies\;&2+a_1+a_2+a_3-a_1a_2a_3 > 0\\[4pt] \implies\;&(2+a_1+a_2+a_3-a_1a_2a_3)a_4 > (2+a_1+a_2+a_3-a_1a_2a_3)(-1)\\[4pt] \implies\;&2a_4+a_1a_4+a_2a_4+a_3a_4-a_1a_2a_3a_4 > -2 -a_1-a_2-a_3 +a_1a_2a_3 \\[4pt] \implies\;&(3+2a_1+2a_2+2a_3+a_1a_2+a_1a_3+a_2a_3)+(2a_4+a_1a_4+a_2a_4+a_3a_4-a_1a_2a_3a_4)\\[4pt] &> \\[4pt] &(3+2a_1+2a_2+2a_3+a_1a_2+a_1a_3+a_2a_3)+(-2-a_1-a_2-a_3+a_1a_2a_3)\\[4pt] \implies\;& 3 + 2(a_1+a_2+a_3+a_4) + a_1a_2+a_1a_3+a_1a_4+a_2a_3+a_2a_4+a_3a_4 - a_1a_2a_3a_4 \\[4pt] &> \\[4pt] &1 + a_1+a_2+a_3 + a_1a_2+a_2a_3+a_3a_1 + a_1a_2a_3 \\[4pt] \implies\;&g(-2) > (1+a_1)(1+a_2)(1+a_3)\\[4pt] \implies\;&g(-2) > 0\\[4pt] \end{align*} Similarly, for $g(2)$, we have the identity $$ g(2) = 3 - 2(a_1+a_2+a_3+a_4) + a_1a_2+a_1a_3+a_1a_4+a_2a_3+a_2a_4+a_3a_4 - a_1a_2a_3a_4 $$ which is our target for the $\text{LHS}$ of the chain below . . . \begin{align*} &1-a_1a_2 > 0\\[4pt] \implies\;&(1-a_1a_2)a_3 < (1-a_1a_2)(1)\\[4pt] \implies\;&a_3-a_1a_2a_3 < 1-a_1a_2\\[4pt] \implies\;&(-2+a_1+a_2)+(a_3-a_1a_2a_3) < (-2+a_1+a_2)+(1-a_1a_2)\\[4pt] \implies\;&-2+a_1+a_2+a_3-a_1a_2a_3 < -1+a_1+a_2-a_1a_2\\[4pt] \implies\;&-2+a_1+a_2+a_3-a_1a_2a_3 < -(1-a_1)(1-a_2)\\[4pt] \implies\;&-2+a_1+a_2+a_3-a_1a_2a_3 < 0\\[4pt] \implies\;&(-2+a_1+a_2+a_3-a_1a_2a_3)a_4 > (-2+a_1+a_2+a_3-a_1a_2a_3)(1)\\[4pt] \implies\;&-2a_4+a_1a_4+a_2a_4+a_3a_4-a_1a_2a_3a_4 > -2 +a_1+a_2+a_3 -a_1a_2a_3 \\[4pt] \implies\;&(3-2a_1-2a_2-2a_3+a_1a_2+a_1a_3+a_2a_3)+(-2a_4+a_1a_4+a_2a_4+a_3a_4-a_1a_2a_3a_4)\\[4pt] &> \\[4pt] &(3-2a_1-2a_2-2a_3+a_1a_2+a_1a_3+a_2a_3)+(-2+a_1+a_2+a_3-a_1a_2a_3)\\[4pt] \implies\;& 3 - 2(a_1+a_2+a_3+a_4) + a_1a_2+a_1a_3+a_1a_4+a_2a_3+a_2a_4+a_3a_4 - a_1a_2a_3a_4 \\[4pt] &> \\[4pt] &1 -a_1-a_2-a_3 +a_1a_2+a_2a_3+a_3a_1 -a_1a_2a_3 \\[4pt] \implies\;&g(2) > (1-a_1)(1-a_2)(1-a_3)\\[4pt] \implies\;&g(2) > 0\\[4pt] \end{align*}

This completes the analysis for $g$.

Answer 2

Here is a partial answer for the first one. Write $$f(x)=x^2-(a_1+a_2+a_3)x+a_1a_2+a_1a_3+a_2a_3-1\ ;$$ I shall show that any real root is less than $2$. The curve $y=f(x)$ is a parabola with vertex at $\frac12(a_1+a_2+a_3)$, which is between $-\frac32$ and $\frac32$; so $f$ is increasing on $[2,\infty)$, and it suffices to show that $f(2)>0$. We have $$f(2)=3-2(a_1+a_2+a_3)+(a_1a_2+a_1a_3+a_2a_3)=g'(1)\ ,$$ where $$g(x)=x^3-(a_1+a_2+a_3)x^2+(a_1a_2+a_1a_3+a_2a_3)x-a_1a_2a_3\ .$$ But $$g(x)=(x-a_1)(x-a_2)(x-a_3)$$ is a cubic with positive leading coefficient and all roots in $(-1,1)$; so $g$ is increasing whenever $x>\max(a_1,a_2,a_3)$, and in particular $g'(1)>0$.