Show that $P$ is divided to simple roots knowing that $a_{k}^2-4a_{k-1}a_{k+1}>0$ 
Let $P(X)=a_0+a_1X+..+a_nX^n\in R[X]$ 
Assume that  $\forall k, a_k>0$ and $a_{k}^2-4a_{k-1}a_{k+1}>0$
Show that $P$  is divided to simple roots in $R[X]$. i.e. $P(X)=C(X−\alpha_1)\cdots(x−\alpha_n)$, where $\alpha_i, \quad 1 \le i \leq n$ are distinct.

I'm stuck wigh this problem, I cannot get any good idea for this.
I tried to use the fact that, 
$$
\sigma_{k}=(-1)^{k}\cdot\frac{a_{n-k}}{a_{n}}
$$
where $\sigma_{k}$ are The elementary symmetric polynomials
But it did not lead nowhere.
In fact I do not see how to connect the fact that $a_{k}^2-4a_{k-1}a_{k+1}>0$ and the result I have to prove.
 A: The assumption can be slightly generalized to $a_k>0$ and
$$
a_k^2> q^2\,a_{k-1}a_{k+1}.
$$
for all $k=1,...,d-1$ for some $q\ge2$.
Because these are finitely many conditions on the coefficients, there also exist $q>2$ for the given set of coefficients.
Define $b_k=q^{k^2}\,a_k$, then 
$$
b_{k-1}b_{k+1}=q^{2k^2+2}\,a_{k-1}a_{k+1}< q^{2k^2}\,a_k^2=b_k^2
$$
Rewriting that relation as
$$
\frac{b_{k-1}}{b_k}<\frac{b_k}{b_{k+1}}
$$
one recognizes that the sequence of these fractions is strictly monotonically increasing.
Now fix some $m$ and consider values 
$$
q^{2m}\,\frac{b_{m-1}}{b_m}=\frac{qa_{m-1}}{a_m}< x< \frac{a_m}{qa_{m+1}}=q^{2m}\,\frac{b_m}{b_{m+1}}
$$

Proposition 1: For these short intervals, $p(-x)$ is non-zero and has the sign $(-1)^m$.

Consider the quotient 
$$
\frac{p(-x)}{(-x)^m}=\dots+(-a_{m-3}\,x^{-3}+a_{m-2}\,x^{-2})+(-a_{m-1}\,x^{-1}+a_m-a_{m+1}\,x)+(a_{m+2}\,x^2-a_{m+3}\,x^3)+...
$$
The claim of the proposition is proved if this fraction is positive.

Lemma 2: All these groups of terms on the right of the fraction are positive.

Check first the group at the center around $a_m$, then the groups to the left and finally to the right. If at one end the group is incomplete, because one of $m$ or $\deg p-m$ is even, then the only term in this group of one is positive.
\begin{align}
-a_{m-1}\,x^{-1}+a_m-a_{m+1}\,x
&> a_m\,(-q^{-1}+1-q^{-1})
\\
&=
a_m\, q^{-1}\,(q-2)
\\&
\ge0
\\[1.2em]%\hline
(-a_{m-2p-1}\,x^{-2p-1}+a_{m-2p}\,x^{-2p})
&=a_{m-2p}\,x^{-2p-1}\,\left(x-\frac{a_{m-2p-1}}{a_{m-2p}}\right)
\\
&> 
a_{m-2p}\,x^{-2p-1}\,\left(q^{2m\,}\frac{b_{m-1}}{b_m}-q^{2m-4p-1}\,\frac{b_{m-2p-1}}{b_{m-2p}}\right)
\\
&>
a_{m-2p}\,x^{-2p-1}\,q^{2m}\,\left(1-q^{-4p-1}\right)\,\frac{b_{m-1}}{b_{m}}
\\&> 0
\\[1.2em]%\hline
a_{m+2p}\,x^{2p}-a_{m+2p+1}\,x^{2p+1}
&=
a_{m+2p+1}\,x^{2p}\,\left(\frac{a_{m+2p}}{a_{m+2p+1}}-x\right)
\\
&>
a_{m+2p+1}\,x^{2p}\,q^{2m}\,\left(q^{4p+1}-1\right)\,\frac{b_{m}}{b_{m+1}}
\\&> 0
\end{align}
One sees that in the sum there is a lower positive bound on $\frac{p(-x)}{(-x)^m}$ for the selected interval.
Conclusion: Since $p(0)=a_0>0$ and the sign of $p(-x)$ for very large $x$ is $(-1)^{\deg p}$, one identifies $\deg p$ disjoint intervals
$$
\left(-\infty, -\frac{a_{\deg p-1}}{qa_{\deg p}}\right],\;
\left[-\frac{qa_{\deg p-2}}{a_{\deg p-1}},-\frac{a_{\deg p-2}}{qa_{\deg p-1}}\right],\;...,\;
\left[-\frac{qa_{1}}{a_{2}},-\frac{a_{1}}{qa_{2}}\right],\;
\left[-\frac{qa_{0}}{a_{1}},0\right]
$$
on the negative ray where $p$ changes its sign and thus must have a real root in between. This already is the number of all roots of $p$, so that the factorization claim follows.

Remark: This question is closely related to the Newton polygon of the polynomial. Indeed, the assumption leads to the equivalent formulation that
$$
-\ln(a_k)=\alpha \,k^2+\phi(k)
$$
where $\alpha=\ln q$ and $\phi$ is a convex function on the integers. The condition on $x$ is related to the slopes of supporting linear functions at the corner $(m,\phi(m))$ of the graph of $\phi$.
Some of that is discussed in the articles of Malajovich/Zubelli on the geometry of the Graeffe iteration resp. the tangent Graeffe iteration.
A: Let $a_0=1$ and define (Following Hardy)
$$a_n=\frac{1}{B_1B_2\cdots B_n},B_k<B_{k+1}$$
Huchinson [1] showed if $a_{k}^2-4a_{k-1}a_{k+1}>0$, then that there exists a simple zero of the polynomial in the interval
$$\left(-\sqrt{B_{2\nu}B_{2\nu+1}},-\sqrt{B_{2\nu-1}B_{2\nu}}\right),\qquad \nu=1,2,...$$
Please refer to [1] for details.
[1] Hutchinson, J. I. "On a remarkable class of entire functions." Transactions of the American Mathematical Society 25.3 (1923): 325-332.
