Problem/Conjecture :
Let $a_i>0$ such that $a_{n}\geq a_{n-1}\geq \cdots \geq a_1\geq a_0$ then define :
$$\sum_{i=1}^{n}a_ix^{2i}+a_0=0$$
Then it seems we have all the complex roots in the unit circle or let $x_{root}=a+bi$ so :
$$a^2+b^2\leq 1$$
I made several random test with Wolfram alpha and it seems true notably as the $a_i\simeq 1$
Motivation :
See https://en.wikipedia.org/wiki/Sendov%27s_conjecture .
For an attempt see my answer here Prove magnitude of zeros of polynomial $\le\frac{\sqrt5+1}2$
On the other hand using Lagrange's bound it's true for :
$$\max\left(1,\sum_{i=0}^{n-1}\frac{a_{i}}{a_{n}}\right)\leq 1$$
But it's the simple case .
Question :
How to (dis)prove it ?(answered now)
The real case :
Let $a_i>0,a_0=-1,n\geq 3$ then define :
$$\sum_{i=1}^{n}a_ix^{i}+a_0=0\tag{I}$$
Then there exits a constant $0<C_n<1/2$ depending on the degree of the polynomial such that :
$$\prod_{i=1}^{n}a_i>C_n>0$$
Then $\exists! x_{root}\in [0,1]$ . Which is a root of $I$
In my answer I provide a lower bound near $C_3\simeq 0.039$ then next bound is lower than $$C_4\simeq 0.0079$$
Have you an asymptotics for $C_n$ ?
To concentrate the effort we can try to show :
$$\frac{C^2_{n-1}}{C_{n}C_{n-2}}\simeq \phi$$
Which is a Turan-Type estimation . Then use https://mathoverflow.net/questions/102540/product-of-fibonacci-numbers
Nota bene :
An very accurate solution (the bound $x=1$ is reached at a limit ) can be found if we consider the equality case in HM-GM considering the coefficients of the highest degree of the polynomials (numerator/denominator in the algorithm) and solving a trinomial using Lagrange reversion theorem from the geometric series (equality case).
Example :
$$\frac{x}{8}(x+1+\frac{y}{x+1})-1=0$$
Now $x\to 2x,y=11$ we have the following iteration :
$$f(x)=\frac{4}{2x+1+\frac{11}{2x+1}}$$
With five iteration with itself ($f(x)$) :
We get the coefficient :
$$\frac{3573\cdot7355}{48631411}\simeq x_{root}$$
Using a proportional argument we get in the case above a solution close to one .
Best Answer
Let us simplify the problem first. For each nonnegative integer $k\le n$ put $b_k=\frac {a_k}{a_n}$. Then $1=b_n\ge\dots\ge b_0$. Put $p(y)=\sum_{k=0}^n b_kx^k$. Then the conjecture is that each root $y$ of $p(y)$ has the absolute value at most $1$. Consider cases of small degree $n$.
Suppose that $n=1$. Then $y=-b_0$ and $|y|=|-b_0|\le 1$.
Suppose that $n=2$. Then $y=-\frac{b_1}2\pm\sqrt{\frac{b_1^2}4-b_0}$. If $\frac{b_1^2}4-b_0\ge 0$ then $y$ is real and $|y|\le \frac{b_1}2+\frac{b_1}2\le 1$. If $\frac{b_1^2}4-b_0<0$ then $y$ is not real and $$|y|^2=\left(\frac{b_1}2\right)^2+\left(b_0-\frac{b_1^2}4\right)=b_0\le 1.$$
Suppose that $n=3$. To be continued...
ErikSatie part:
As quoted by user MartinR see Showing that the roots of a polynomial with descending positive coefficients lie in the unit disc.