Let \[ 0<a_0\le a_1\le\cdots\le a_n. \] Prove that \[ a_0z^n+a_1z^{n-1}+\cdots+a_n=0 \] has no roots in the disk \(|z|<1\).
Solution
Put \[ P(w)=a_0w^n+a_1w^{n-1}+\cdots+a_n. \]
1. If \(P(z)=0\) and \(\abs{z}<1\), then\[ 0<\abs{z}<1. \]
Since \(a_n>0\), \[ P(0)=a_n\neq0. \] Thus a root \(z\) cannot equal zero.
2. If \(P(z)=0\), then\[ a_n = a_0z^{n+1} + \sum_{j=1}^n (a_j-a_{j-1})z^{n-j+1}. \]
Direct expansion gives \[ \begin{aligned} (z-1)P(z) &= a_0z^{n+1} + (a_1-a_0)z^n + \cdots\\ &\qquad + (a_n-a_{n-1})z - a_n. \end{aligned} \] If \(P(z)=0\), the left-hand side is zero. Rearranging gives the claimed identity.
3. A root \(z\) with \(\abs{z}<1\) would satisfy\[ a_n<a_n, \]which is impossible.
Let \[ r=\abs{z}. \] By step 1, \(0<r<1\). Since \[ a_j-a_{j-1}\geq0 \] for \(1\leq j\leq n\), step 2 and the triangle inequality give \[ \begin{aligned} a_n &\leq a_0r^{n+1} + \sum_{j=1}^n (a_j-a_{j-1})r^{n-j+1}\\ &\leq r \left( a_0 + \sum_{j=1}^n(a_j-a_{j-1}) \right)\\ &= ra_n\\ &< a_n. \end{aligned} \] The penultimate equality is the telescoping identity \[ a_0+\sum_{j=1}^n(a_j-a_{j-1})=a_n, \] and the last inequality uses \(a_n>0\) and \(r<1\).
4. Therefore\[ \boxed{P(z)\neq0\quad\text{whenever }\abs{z}<1}. \]
If such a root existed, step 3 would give a contradiction.
5. Q.E.D.
Step 4 is the required conclusion.