Proving that $\sum\limits_{n = 0}^{2013} a_n z^n \neq 0$ if $a_0 > a_1 > \dots > a_{2013} > 0$ and $|z| \leq 1$ I'm going to teach a preparation course for the complex analysis qualifying exam from my university (which basically consists of me doing some problems from past exams) and I'm trying to solve some questions from previous exams. One of the questions is the following: 


Prove that $\displaystyle{\sum_{n = 0}^{2013} a_n z^n \neq 0}$ if the coefficients  satisfy $a_0 > a_1 > \dots > a_{2013} > 0$ and $|z| \leq 1$.


I tried to approach the problem by using the reverse triangle inequality to try to isolate the leading term, which I thought should be the constant coefficient, as follows:
\begin{array}
. \left|  \sum_{n = 0}^{2013} a_n z^n \right| &= \left| a_0 +  \sum_{n = 1}^{2013} a_n z^n \right| \\
&\geq |a_0| - \left|  \sum_{n = 1}^{2013} a_n z^n \right|\\
&\geq |a_0| -  \sum_{n = 1}^{2013} |a_n z^n|\\
&\geq a_0 -  \sum_{n = 1}^{2013} a_n |z^n|\\
&\geq a_0 -  \sum_{n = 1}^{2013} a_0 |z^n|\\
&= a_0\left( 1  -  \sum_{n = 1}^{2013} |z^n| \right)
\end{array}
Then for $|z| < 1$, we have that this last expression is equal to
$$
a_0\left( 1  -  |z| \frac{|z|^{2013} - 1}{|z| - 1} \right)
$$
and my hope was to prove that this was always positive, nevertheless, after failing to prove it, I plotted the expression inside the parenthesis as a function of $|z|$, and found out that it changes sign in the interval $(0, 1)$.
Question
Can my approach somehow be made to work? And if not, how can I prove that the sum $\displaystyle{\sum_{n = 0}^{2013} a_n z^n \neq 0}$ under the given conditions?
Thank you very much for any help. 
 A: You can look at (with $N$ = 2013).
$$
g(z) = zf(z) - f(z) = a_N z^{N+1} + (a_{N-1} - a_N) z^{N} + (a_{N-2} - a_{N-1}) z^{N-1} + \cdots + (a_0 - a_1) z - a_0
$$
Since $g(z) = (z-1)f(z)$, strictly inside the unit disk the zeros of $g$ are the same as the zeros of $f$. On the boundary $g$ has a zero at $z = 1$, which needs to be handled separately.
$$
|g(z)| \ge \Big| |a_0| - |a_N z^{N+1} + (a_{N-1} - a_N) z^{N} + (a_{N-2} - a_{N-1}) z^{N-1} + \cdots + (a_0 - a_1) z | \Big|
$$
$$
|g(z)| \ge \Big| |a_0| - |a_N + (a_{N-1} - a_N)  + (a_{N-2} - a_{N-1}) + \cdots + (a_0 - a_1)  | \Big|
$$
$$
|g(z)| \ge \Big| |a_0| - |a_0 | \Big| = 0
$$
Strictly inside the unit disk there is never equality. When $|z| = 1$ you get equality when $z, z^2, \cdots z^{N+1}$ are collinear, which only happens at $z=1$. But $z = 1$ is not a zero of $f$.
A: Your approach cannot work because if e.g. $a_0\to-a_0$ the result stops working, but your argument essentially depends only on $|a_0|$.
As pointed out in the comments, this result is shown elsewhere, such as at Showing that the roots of a polynomial with descending positive coefficients lie in the unit disc.

Expanding on the above, let $a_0=1,a_1=\frac 3 4,a_2=\frac 1 2$ and everything else vanish (or be $\epsilon$, whatever). Then the first inequality you write down has
$$|a_0|-\left|\sum_{n=1} a_nz^n\right| \text{ which at }z=1\text{ is just } 1-\frac 5 4 - \cdots \le -\frac 1 4$$
so you have no hope of bounding it below by zero. You need to use the relationship between $a_0$ and the other $a_i$, you cannot simply split it off via a single application of the triangle inequality.
A: Instead of an answer I will state a slightly stronger assertion:
$\newcommand{\CC}{\mathbb{C}}$

Let $a_0\geq a_1\geq\cdots\geq a_n>0$, $n\geq 1$,
  and set $f(z):=\sum_{k=0}^na_kz^k~$ for $z\in\CC$.
  If $a_{k-1}>a_k$ for some integer $k$, $1\leq k\leq n$, that is coprime to $n+1$
  (in particular if $a_0>a_1$ or $a_{n-1}>a_n$),
  then $f(z)\neq 0$ whenever $|z|\leq 1$.

See my answer to this
duplicate question.
Here is a more precise result:
$\newcommand{\set}[1]{{\{#1\}}}
$$\newcommand{\suchthat}{\mid}
$$\newcommand{\union}{\cup}
$$\newcommand{\NN}{\mathbb{N}}
$

Let $a_0\geq a_1\geq\cdots\geq a_n>0$, $n\geq 1$,
  and set $f(z):=\sum_{k=0}^na_kz^k~$ for $z\in\mathbb{C}$,
  $K:=\set{k\in\NN\suchthat\text{$1\leq k\leq n$ and $a_{k-1}>a_k$}}$,
  $d:=\gcd(K\union\set{n\!+\!1})$.
  $~$If $d=1$, then $f(z)\neq 0$
  for every $z$ in the closed unit disk $D:=\set{z\in\CC\suchthat |z|\leq 1}$.
  $~$If $d>1$, then all the zeros of $f(z)$ that lie in $D$
  are $\,\exp(l\!\cdot\!2\pi i/d)$, $1\leq l\leq d-1$,
  and every one of them is simple.

