Let $P(z)=a_0+a_1z+\cdots+a_nz^n$ be a polynomial whose coefficients satisfy $$0<a_0<a_1<\cdots<a_n.$$
I want to show that the roots of $P$ live in unit disc. The obvious idea is to use Rouche's theorem, but that doesn't quite work here, at least with the choice $f(z)=a_nz^n, g(z)=$ (the rest).
Any ideas?
The thing to do is to look instead at the polynomial $$Q(z) = (1-z)P(z) = (1-z)\left(\sum_{i=0}^n a_iz^i \right) = a_0 -a_n z^{n+1} + \sum_{i=1}^n (a_i-a_{i-1})z^i$$ Now, let $|z|>1$ be a root of $P(z)$, and hence a root of $Q(z)$. Therefore, we have $a_0 + \sum_{i=1}^n (a_i-a_{i-1})z^i = a_n z^{n+1}$ Then, we have \begin{aligned} |a_n z^{n+1}| &= \left|a_0 + \sum_{i=1}^n (a_i-a_{i-1})z^i\right| \\ & \le a_0 + \sum_{i=1}^n (a_i-a_{i-1})|z^i| \\ & < a_0|z^n| + \sum_{i=1}^n (a_i-a_{i-1})|z^n| \\ & = |a_n z^n|\end{aligned} a contradiction.
For a nice article on integer polynomials, see here. (Your problem is Proposition 10)
The previous answer establishes that roots lie inside or on the unit circle. By a slight modification to $$ Q(z)=(1-z)P(Rz) $$ this argument remains true as long as $0<a_0\le Ra_1 \le R^2a_2\le ... \le R^na_n$. Let $R$ be minimal with this property, then $R<1$. Now if $z$ is any root of $P$, then $z/R$ is a root of $Q$ and thus in the unit disk. Thus $|z|\le R<1$.
A very classical proof
Similar idea to other but slightly different point of view (symetric polynomial):
I let the proof as an exercise :
Considering the geometric series $$Q(x)=\sum_{i=0}^1x^i/a^i,a\in(0,1]$$ show the roots are in the unit circle .
Show that every polynomial :$P(x)$ (degree $n$ with all real roots) with decreasing coefficient and positive can be written as the product of polynomials $Q_i(x)=a_ix+1,a_i>1$ with $C>0$ a positive constant or :
$$P(x)=C\prod_{i=1}^{n}Q_i(x)$$
Hint use fundamental theorem of algebra.
As the propotionality doesn't affect the origin of the roots conclude .
Generalize with $ax^2+bx+c,a>b>c>0$
