If $a_0, a_1, \ldots a_{n-1}$ are positive real numbers, then all zeros of the polynomial $p(z)=a_0z^{n-1}+a_1z^{n-2}+\ldots + a_{n-1}$ lie in the annulus $A(0,m,M)$ where $m,M$ are the minimum and maximum of ratios of the coefficients $a_i$
I have recently learned Rouche's so I think I just have to find a clever function $q$ and compare it with $p$. I understood the proof of Enestrom-Kakeya given here but am unsure how to proceed to handle the general case.
Let $c$ maximum of the ratio of the coefficients and consider $$q(z)=(1-z)p(cz)$$ so
$$q(z)=a_{n-1}+\sum_{k=1}^{n-1}(c^ka_{n-1-k}-c^{k-1}a_{n-k})z^k-a_0c^{n-1}z^n$$
We notice that if we write $q(z)=\sum_{k=0}^n b_kz^k$, we have $b_k \ge 0, k=0,...n-1$ by our choice of $c$ so by telescoping (or simply noting that $\sum_{k=0}^n b_k=0$ as $q(1)=0$) we have that $$\sum_{k=0}^{n-1}|b_k/b_n|=1$$ hence the usual Lagrange bound for the roots of $q$, gives that the highest such has absolute value at most $\max (1, \sum_{k=0}^{n-1}|b_k/b_n|)=1$
But any root $w$ of $p$ gives a root $w/c$ of $q$, so we get that the highest root of $p$ has absolute value at most $c$ as required.
Applying the above result to the polynomial $p^*(z)=z^{n-1}p(1/z)=\sum_{k=0}^{n-1}a_kz^k$ clearly gives the result for the minimum value of the absolute value of the roots of $p$, so we are done!
