I am trying to solve the following problem: Consider $n\geq 2$ and the polynomial $$P(x)=x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_1x+a_0$$
with $a_{n-1}, a_{n-2},\ldots,a_1, a_0\in[0,1]$. Prove that if $P(z)=0$, then $\Re(z)<\dfrac{1+\sqrt{5}}{2}$. ($\Re(z)$ denotes the real part of the complex number $z$).
I can't seem to find a meaningful way to use the condition $a_i\in[0,1]$. I managed to prove that if $P(z)=0$, then $|z|<2$:
$$|z|^n=|a_{n-1}z^{n-1}+a_{n-2}z^{n-2}+\ldots+a_1z+a_0|\leq\sum_{k=1}^{n-1}|z|^k$$
hence, if $|z|>1$,
$$|z|^n\leq\frac{|z|^n-1}{|z|-1}\Rightarrow |z|^{n+1}\leq 2|z|^n-1\Rightarrow |z|<2-\frac{1}{|z|^n}$$
and so, $|z|<2$.
Also, we see here another bound for $|z|$ in terms of the positive coefficients, but, nonetheless, I couldn't find anything about $\Re(z)$.
Best Answer
Let $g=\frac{1+\sqrt 5}{2}$ be the golden ratio. Note that for $n \ge 2$ one has $g^n (g-1)=g^{n-1}>g^{n-1}-1$ so $g^n >1+g+\ldots +g^{n-2}$
But now assume by contradiction that the equation above has a root $\Re z \ge g$; dividing by $z^{n-1}$ one has:
$$ |z+a_{n-1}|=|-a_{n-2}/z-\ldots-a_0/z^{n-1}| \le 1/|z|+\ldots +1/|z^{n-1}|$$
But now $\Re(z+a_{n-1}) \ge \Re z >0$ and the imaginary parts are same so $|z| \le |z+a_{n-1}|$ so by substituting and simplifying we get
$$|z|^n \le 1+|z|+\ldots+|z|^{n-2}$$
But now the equation $x^n=1+\ldots+x^{n-2}$ has only one positive root $r_n$ (division by $x^n$ shows that clearly as rhs decreases then strictly), and from the left one has $0<x<r_n$ implies $x^n <1+x+\ldots +x^{n-2}$ (while from the right we have the opposite inequality). So, $r_n <g \le |z|$ on one hand by the first paragraph, and $|z| \le r_n$ by the inequality above which is a contradiction, and we are done!