A pen-and-paper proof for a matrix implication. Suppose $A = \begin{bmatrix} x & 1\\ y & 0\end{bmatrix}, B = \begin{bmatrix} z & 1\\ w & 0\end{bmatrix}$, for $x,y,z,w \in \Bbb{R}$.
I have observed by considering many  examples of $x,y,z,w$ that:
If all the eigen values of $A^2B$ and $AB^2$ are less than one in absolute value $\implies$ $\det(AB+A+I)<0$ and $\det(BA+B+I)<0$ is not possible.
OR alternatively,
If all the eigen values of $A^2B$ and $AB^2$ are less than one in absolute value $\implies$ $\det(AB+A+I)\ge 0$ OR $\det(BA+B+I)\ge 0$
I wonder how to prove it actually?
A computational proof using computer package was shown in
https://mathoverflow.net/questions/435267/proof-of-a-matrix-implication/435689#435689
But I am wondering about formal or analytical proof for this question which can be done using pen paper.
$\textbf{EDIT}$
$\textbf{The case of $y=x$ or $z=w$ is covered by Andreas as an answer below.}$
$\textbf{The only case that remains to show is whether the conjecture still holds for $y\neq x$ or $w \neq z$.}$
$\textbf{The bounty is for this purpose if some has ideas on it?}$
 A: @all who want to solve the riddle or find a counterexample:
If I made no mistake, then the eigenvalues are
\begin{align*}
2\lambda_1&=\alpha+ \sqrt{\alpha^2+4xy\alpha + 4y^2(w-x^2)}\, , \,\alpha=(x^2+y)z+x(y+w)\\
2\lambda_2&=\alpha- \sqrt{\alpha^2+4xy\alpha + 4y^2(w-x^2)}\\
2\lambda_3&=\beta +\sqrt{\beta^2+4yz\beta +4y(w^2-yz^2)}\, , \,\beta=x(z^2+w)+(y+w)z\\
2\lambda_4&=\beta -\sqrt{\beta^2+4yz\beta +4y(w^2-yz^2)}
\end{align*}
and the determinants are
\begin{align*}
\delta_1:=\det(AB+A+I)&=(x-y)z+(y+1)w+(x+1)\\
\delta_2:=\det(BA+B+I)&=(z-w)x+(w+1)y+(z+1)
\end{align*}
The statement now reads:
If all $|2\lambda_i|<2$ for $i=1,2,3,4$ then $\delta_1>0$ or $\delta_2>0.$
And to save time: $w=0\wedge z=-y$ has no counterexample according to WA.
A: The claim in the question  can be rephrased by taking the inverse formulation:
$\det(AB+A+I) < 0 \qquad $ AND $\qquad \det(BA+B+I) < 0$
$\implies$ At least one eigenvalue of $(A^2B$ ; $AB^2)$ is greater or equal than one in absolute value.
As a boundary case, there are indeed situations with
$\det(AB+A+I) = 0$ ;  $\det(BA+B+I)= 0 $    ;
where at least one eigenvalue of $(A^2B$ ; $AB^2)$ equals $1 +q$ where $q > 0$.
This is e.g. achieved with values  $x = w = z = -1 $ and $y = -1-q$, then we have
$$
\det(AB+A+I) = (x-y)z+(y+1)w+(x+1)  = 0\\
\det(BA+B+I)=(z-w)x+(w+1)y+(z+1) = 0 
$$
and the characteristic equations for the eigenvalues read
$$
\det(A^2 B-\lambda I) = (z(x^2+y)+xw - \lambda)(x y - \lambda) -  (x^2+y)(xyz+wy) = (1+q -\lambda )^2 = 0 \\
\det(A B^2-\lambda I) = (x(w+z^2)+wz - \lambda) (y z - \lambda) - (xyz+wy)(z^2+w) = (1+q -\lambda )(1 -\lambda )= 0
$$
Hence we have eigenvalues $\lambda = 1+q > 1$ for both $A^2 B$ and $A B^2$.
This could also support possible contradictions to the claim, however not sufficiently. It gives rise to be able, by a continuity argument, to transport this to the  adjacent domain $\det(AB+A+I) < 0 \qquad $ AND $\qquad \det(BA+B+I) < 0$. We can look at this domain by observing
$$
\det(AB+A+I) = (x-y)z+(y+1)w+(x+1)  = (y-x)(w-z) + (x+1)(w+1) \tag{1}\\
\det(BA+B+I)=(z-w)x+(w+1)y+(z+1) =  (y-x)(w-z) + (y+1)(z+1)
$$
Hence a particularly convenient subcase is the one with $y=x$ and $w=z$. Here,  $\det(AB+A+I) < 0$ is achieved in the two cases $(x>-1 ; z < -1)$ and $(x<-1 ; z > -1)$. It is only necessary to consider one of these cases, since exchanging  $x \leftrightarrow z$ exchanges  $A \leftrightarrow B$ and this is covered by the second determinant inequality. As we are only aiming at showing that one eigenvalue will always be  greater or equal than one in absolute value, it suffices to establish one case.
We have the characteristic equation for $A^2 B$:
\begin{align}
& \det(A^2 B  - \lambda I) =
 \lambda^2 - \lambda(x z^2 + z^2 + 2 x z) - x z^2 = 0 
\end{align}
Now let $x =  -1 +a$ and $z = -1 -b$ with $a,b > 0$, then we get
$$
(\lambda-1)^2 + \lambda(-a b^2- 2 b +a) -(a - 2 b + 2a b + b^2 a - b^2) = 0
$$
which can be solved for $\lambda$. To simplify things, it suffices to notice that the equation
$$
(\lambda-1)^2 + \lambda q - (q + r) = 0
$$
has the largest solution
$$
\lambda_1 = 1 - q/2 + \frac12 \sqrt{q^2 + 4r} 
$$
Hence we have, if $r >0$, that $\lambda_1 > 1 +  (1 - {\rm sign}(q)) \; |q|/2 > 1$. Applied to the characteristic equation, this is
$$
r = (a - 2 b + 2a b + b^2 a - b^2)- (-a b^2- 2 b +a)  = b(2 a b + 2 a  - b) 
$$
hence if $2 a b + 2 a  - b > 0$ then $A^2 B$  has an eigenvalue $\lambda > 1$.
Likewise, the characteristic equation for $A B^2$ gives that if $2b - a - 2ab>0$, then $A B^2$  has an eigenvalue $\lambda > 1$.
Adding the two conditions gives $a+b$ which is $> 0$, hence it is impossible that both conditions fail (i.e. both are negative). Hence, at least one eigenvalue  of $(A^2B$ ; $AB^2)$ is greater or equal than one in absolute value.
This establishes the claim for the  subcase  $y=x$ and $w=z$. The next step is to exploit the  determinant equations (1) further for the case $y\ne x$ or $w\ne z$. [...]
