Here, we shall use methods similar to those in the answer by Gabriel Romon to rule out the case of degree $5$ polynomials. We start by sharpening some of the bounds from that answer and then we introduce the idea that evaluating a polynomial at the value $1$ is the same as summing its coefficients. Enjoy!
Let $x^r+a_1x^{r-1}+a_2x^{r-2}+\dots+a_r$ be a polynomial with nonzero real coefficients such that $a_1,\dots,a_r$ are all roots (counting multiplicity) so that
$$(x-a_1)\cdots(x-a_r) = x^r+a_1x^{r-1}+a_2x^{r-2}+\dots+a_r$$
(Note: We shall never consider the case $r\le 2$ below.) Equating coefficents, we have $a_1 = -\sum_{i=1}^ra_i$, as well as $a_2=\sum_{1\le i<j\le r}a_ia_j$, and $a_r=(-1)^r\prod_{i=1}^ra_i$.
Thus, $a_1^2 = \sum_{i=1}^ra_i^2 + 2a_2$ so that $0 = a_2^2+2a_2+\sum_{i=3}^ra_i^2$. The discriminant of this quadratic polynomial is nonnegative and the roots of the given polynomial are nonzero, so $0<\sum_{i=3}^ra_i^2\le 1$. In particular, $a_2^2+2a_2<0$. This means $a_2\in(-2,0)$.
Having reiterated what we need from an earlier post, let's get to work.
PRP 1:
If $r\ge 4$, then
$|a_1|\le \frac{2+\sqrt{r-2}}2$.
$\prod_{j=3}^r|a_j| \le \left(\frac{\sqrt{-a_2}\sqrt{2+a_2}}{\sqrt{r-2}}\right)^{r-2}$.
if $3\le j_0\le r$ then $\prod_{3\le j\le r\\j\ne j_0}|a_j| < \left(\frac{\sqrt{-a_2}\sqrt{2+a_2}\sqrt{r-2}}{r-3}\right)^{r-3}$.
Proof:
1. Using the Cauchy-Schwarz inequality and $|a_2|\le 2$, we have
\begin{equation*}
|a_1| \le \frac12(|a_2|+\sum_{i=3}^r|a_i|) \le \frac{2+\sqrt{r-2}}2.
\end{equation*}
2-3. We use $\sum_{j=3}^ra_j^2 = -a_2(2+a_2)$. By the Cauchy-Schwarz inequality, we have
\begin{equation*}
\sum_{j=3}^r|a_j| \le \sqrt{-a_2}\sqrt{2+a_2}\sqrt{r-2}.
\end{equation*}
Also, since $a_j\ne 0$ for $j=1,\dots,r$, we get
\begin{equation*}
\sum_{3\le j\le r\\j\ne j_0}|a_j| < \sqrt{-a_2}\sqrt{2+a_2}\sqrt{r-2}
\end{equation*}
Now, we apply the AM-GM inequality to deduce (2.), (3.).
We purposely kept instances of $a_2$ in the bounds in PRP 1. We are going to multiply by $|a_2|$ and take a maximum of a function of $|a_2|$ using the following lemma, which is proved with methods of single-variable calculus.
LEM 2:
If $a,b>0$, then the function $f(x)=x^a(2-x)^b$ has its maximum on $[0,2]$ at $x=2a/(a+b)$.
Here's what we get.
PRP 3:
If $r\ge 4$, then
$\prod_{j=2}^r|a_j| \le \frac{r^{r/2}}{(r-1)^{r-1}}$.
for $3\le j_0\le r$ we have $\prod_{2\le j\le r\\j\ne j_0}|a_j| \le \frac{(r-1)^{(r-1)/2}}{(r-3)^{(r-3)/2}(r-2)^{(r-1)/2}}$.
Proof:
1. By PRP 1.2 and LEM 2, we have
\begin{align*}
\prod_{j=2}^r|a_j| &\le \frac{|a_2|^{r/2}(2-|a_2|)^{(r-2)/2}}{(r-2)^{(r-2)/2}} \\
&\le \frac{(r/(r-1))^{r/2}((r-2)/(r-1))^{(r-2)/2}}{(r-2)^{(r-2)/2}} \\
&= \frac{r^{r/2}}{(r-1)^{r-1}}
\end{align*}
- Using PRP 1.3 and including LEM 2, we derive the second inequality as follows
\begin{align*}
\prod_{2\le j\le r\\j\ne j_0}|a_j| &\le \frac{|a_2|^{(r-1)/2}(2-|a_2|)^{(r-3)/2}(r-2)^{(r-3)/2}}{(r-3)^{r-3}} \\
&\le \frac{((r-1)/(r-2))^{(r-1)/2}((r-3)/(r-2))^{(r-3)/2}(r-2)^{(r-3)/2}}{(r-3)^{r-3}} \\
&=\frac{(r-1)^{(r-1)/2}}{(r-2)^{(r-1)/2}(r-3)^{(r-3)/2}}
\end{align*}
Next, we use the fact that $a_1a_2\dots a_{r-1}=(-1)^r$ to derive some fresh inequalities.
PRP 4:
If $r\ge 4$, then
$\frac{(r-3)^{(r-3)/2}(r-2)^{(r-1)/2}}{(r-1)^{(r-1)/2}}\le |a_1|$.
$|a_r|\le \frac{r^{r/2}}{(r-1)^{r-1}}\cdot\frac{2+\sqrt{r-2}}2$.
$a_2\le-\frac2{2+\sqrt{r-2}}\cdot\left(\frac{r-3}{\sqrt{r-2}}\right)^{r-3}$
Proof:
From $(-1)^r=\prod_{i=1}^{r-1}|a_i|$, we derive
$$|a_1| = \frac 1{\prod_{i=1}^{r-1}|a_i|}.$$
The desired inequality follows from PRP 3.2.
Likewise $\prod_{i=2}^{r-1}|a_i|=1/|a_1|\ge 2/(2+\sqrt{r-2})$ by PRP 1.1. Thus,
$$\frac2{2+\sqrt{r-2}}|a_r| \le \prod_{i=2}^{r}|a_i|.$$
Then, use PRP 3.1.
By PRP 2.3,
$$\prod_{j=3}^{r-1}|a_j| < \left(\frac{\sqrt{-a_2}\sqrt{2+a_2}\sqrt{r-2}}{r-3}\right)^{r-3} \le \left(\frac{\sqrt{r-2}}{r-3}\right)^{r-3}$$
Thus,
$$\frac2{2+\sqrt{r-2}}\le \prod_{j=2}^{r-1}|a_j| \le |a_2|\left(\frac{\sqrt{r-2}}{r-3}\right)^{r-3}$$
as desired.
Case $r=5$
Now, we focus on the case $r=5$. Observe that PRP 4.1 simplifies to $|a_1|\ge 9/8$. In particular $a_1\ne 1$. So, from the equation
$\prod_{i=1}^5(1-a_5) = 1+\sum_{i=1}^5a_i = 1-a_1$,
we get
\begin{equation*}
\frac 1{(1-a_2)(1-a_5)} = (1-a_3)(1-a_4) = 1-a_3-a_4+a_3a_4
\end{equation*}
Using the upper bounds on $a_2$ and $a_5$ from PRP 4, we get
\begin{equation*}
a_3+a_4-a_3a_4 = 1-\frac1{(1-a_2)(1-a_5)} \ge 1-\frac1{(1-0.41)(1-(-0.72))}>0
\end{equation*}
We next find a lower bound for $a_3a_4$. We note that $a_1$ is positive. For if $a_1<0$, then $a_3+a_4+a_5>-a_2+9/4>2.25$. We have a contradiction, since $\sqrt3\ge |a_3|+|a_4|+|a_5|$ by Cauchy-Schwarz. So
\begin{equation*}
a_1 \ge 9/8=1.125.
\end{equation*}
From $-1=a_1a_2a_3a_4$, and using $a_1\le (2+\sqrt(3))/2$ and $-a_2<2$, we find
$$a_3a_4 = \frac1{a_1(-a_2)} > \frac 1{2+\sqrt3} > 0.25$$
Hence,
\begin{equation*}
a_3+a_4 > a_3a_4 > 0.25.
\end{equation*}
Therefore, using $a_2\ge -2$ and $a_5\ge -0.41$,
\begin{equation*}
a_1 = \frac12(-a_2-a_3-a_4-a_5) \le \frac12(2.41-0.25) =1.08.
\end{equation*}
We have a contradiction.
