Let $(a_0,a_2\dots,a_{n-1})\in[0,1]^n$, and $k\in\Bbb C$ be a zero of $f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$. Prove that $|k|\le\frac{\sqrt5+1}2$.
Firstly, $\frac{\sqrt5+1}2$ seems to come from quadratics such as $x^2-x-1=0$. So I tried the triangle inequality to obtain a quadratic inequality about $|k|$: $$\sum_{i=0}^{n-1}a_i|k|^i\ge\left|\sum_{i=0}^{n-2}a_ik^i\right|=\left|k^n+a_{n-1}k^{n-1}\right|=|k|^{n-1}|k+a_{n-1}|.$$ I'm not sure whether this is the right way...
For the real case you have in term of Continued fraction a solution here . Got a factored version of the Taylor's series?
Some details :
Let $a_i>0$ then define :
$$P(x)=\sum_{i=0}^{n}a_ix^{2i}=0$$
Or squaring:
$$(\sum_{i=1}^{n}a_ix^{2i})^2=(a_0)^2$$
Now it preserves one roots which have a pure imaginary part (need an argument I don't know such).
Or :
$$\sum_{i=1}^{n}a_{k}^{2}x^{4k}+2\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}a_{i}a_{j}x^{i+j}=a_{0}^{2}\tag{J}$$
Now we translate this polynomial into :
$$x\left(b_0(1+x)+\frac{b_1}{x+1}+\frac{b_2}{1+\frac{1}{x+1}}+\cdots+\frac{b_{n}}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots x}}}}\right)-C=0$$
Or squaring and $x\to x^2$:
$$Q(x)=x^2\left(b_{0}\left(x^2+1\right)+b_{1}\cdot\frac{1}{x^2+1}+b_{2}\frac{x^2+1}{x+2}+b_{3}\cdot\frac{x^2+2}{2x^2+3}+\cdot\cdot+\frac{b_{n}\left(F_{n-2}x^2+F_{n-1}\right)}{F_{n-1}x^2+F_{n}}\right)^2=C^2 \tag{J}$$
To have the Fibonacci's polynomials put all on one denominator and use
$$b_{k}=\left(-1\right)^{k}a_{k}^{2}+2\sum_{j=0}^{k-1}\left(-1\right)^{j}a_{j}a_{2k-j}$$
Where :
$$p(x)=\sum_{i=0}^{n}a_ix^{i},q(x)=\sum_{i=0}^{n}b_ix^{i}$$
Here $I,J$ .
Now setting $Q(mx),\exists m>1$ such that the coefficient decreases starting from the highest degree .
We have using $HM-GM$ ,$x_{root} \leq 1$
Example :
Let :
$$u(x+1)+\frac{y}{x+1}+\frac{z}{1+\frac{1}{1+x}}$$
Or :
$$\frac{ux^{3}+4ux^{2}+5ux+2u+x^{2}z+xy+2xz+2y+z}{x^{2}+3x+2}$$
Then using HM-GM we have the essential expression :
$$\left(u(x+1)\frac{y}{x+1}\frac{z}{1+\frac{1}{1+x}}\right)^{-1}$$
Iterating one time :
$$\left(zyu(xuyz+x+uyz+2)\left(uxyz+uyz+2x+4\right)^{-1}\right)^{-1}$$
And the conclusion follows from for $uyzx>1$
$$\frac{1}{9}\left(uyz(xuyz+x+uyz+2)\left(uxyz+uyz+2x+4\right)^{-1}\right)^{-\frac{2}{3}}<1$$
Which is obvious .
