Domain over which a polynomial is strictly positive

Question

$$Q(x) = x^k + a_1x^{k+1}+ ... + a_nx^{k+n}$$ where $k,n$ are positive integers is a polynomial with real coefficients. I have to show that $Q(x)/x^k$ is strictly positive for all real x satisfying $$0<|x|<1/(1+\sum_{i=0}^n |a_i|)$$ Source: ISI UGB 2017

Now the polynomial formed after division(which is legal since $x$ is non-zero) is the general $n$ degree polynomial $1+a_1x+...+a_nx^n.$ This approaches $1$ as $x$ approaches $0$. But how can I obtain the right hand part of the inequality? Furthermore, I tried substituting $x=1$ to obtain the right hand denominator but the it doesn't give the sum of the absolute values.

Answer 1

We will show that $$ p(x)=1+a_1x+\ldots+a_nx^n>0\quad\text{for all }x\in X=\left[-\frac{1}{1+\sum|a_i|},\frac{1}{1+\sum|a_i|}\right]. $$ Observations:

1. $|x|\le\frac{1}{1+\sum|a_i|}\le 1$ $\Rightarrow$ $|x|^i\le|x|$.
2. $p(x)=1+\sum a_ix^i\ge 1-\max_{x\in X}|\sum a_ix^i|$.

We can estimate from 1 $$ |\sum a_ix^i|\le\sum|a_i|\underbrace{|x|^i}_{\le |x|}\le|x|\sum|a_i|\le\frac{1}{1+\sum|a_i|}\sum|a_i|<1,\quad\forall x\in X, $$ then $\max_{x\in X}|\sum a_ix^i|<1$ and from 2 we conclude that $p(x)>1-1=0$ in $X$.