Actually, no matter how large $A$ is, there will be nonzero solutions that will converge to zero, so you can't prove unboundedness, even though it may be true that the 'generic' solution is unbounded. Here is why: First, convert the system into a first order system in $\mathbb{R}^3$ by setting $x = x_0$, $\dot x = x_1$, and $\ddot x = x_2$. Then the equation becomes the first order system $$ \begin{aligned} \dot x_0 &= x_1\,,\\ \dot x_1 &= x_2\,,\\ \dot x_2 &= -x_0 + {x_1}^2 - A\,x_2\,. \end{aligned} $$ This vector field (i.e., ODE) has one singular point, $(x_0,x_1,x_2) = (0,0,0)$, and its linearization at this point has the matrix $$ \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ -1 & 0 & -A \end{pmatrix} $$ The eigenvalues of this matrix are the roots of $\lambda^3-A\lambda^2 - 1 = 0$, and, when $A\ge0$, there is one real root $r>0$ and two complex conjugate roots $z$ and $\bar z$. These roots satisfy $z\bar z r = 1$ and $z\bar z + r(z+\bar z) = 0$, so we have $z+\bar z = -1/r^2 < 0$, so the real part of $z$ is negative. Hence, there is a stable manifold of dimension 2 (and an unstable manifold of dimension 1), say $M^2\subset\mathbb{R}^3$ that passes through $(0,0,0)$ and is such that any initial condition starting on $M$ will spiral in to $(0,0,0)$. Thus, such solutions will stay bounded for all time, in fact, they'll decay to zero.

Added remark: In fact, even when $A$ is negative, the roots of $\lambda^3-A\lambda^2 - 1 = 0$ are one positive real and either two complex conjugate roots with negative real part or else two real negative roots. Consequently, the stable manifold through $(0,0,0)$ always has dimension $2$, so there are always decaying solutions that are not identically zero.

Actually, no matter what $A$ is, there will be nonzero solutions that will converge to zero, so you can't prove unboundedness, even though it may be true that the 'generic' solution is unbounded. Here is why: First, convert the system into a first order system in $\mathbb{R}^3$ by setting $x = x_0$, $\dot x = x_1$, and $\ddot x = x_2$. Then the equation becomes the first order system $$ \begin{aligned} \dot x_0 &= x_1\,,\\ \dot x_1 &= x_2\,,\\ \dot x_2 &= -x_0 + {x_1}^2 - A\,x_2\,. \end{aligned} $$ This vector field (i.e., ODE) has one singular point, $(x_0,x_1,x_2) = (0,0,0)$, and its linearization at this point has the matrix $$ \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ -1 & 0 & -A \end{pmatrix} $$ The eigenvalues of this matrix are the roots of $\lambda^3+A\lambda^2 + 1 = 0$, and there is always at least one negative real root. Hence the stable manifold of $(0,0,0)$ has dimension at least $1$, so there will always be a nonzero solution that decays to zero (in infinite time).

Actually, no matter how large $A$ is, there will be nonzero solutions that will converge to zero, so you can't prove unboundedness, even though it may be true that the 'generic' solution is unbounded. Here is why: First, convert the system into a first order system in $\mathbb{R}^3$ by setting $x = x_0$, $\dot x = x_1$, and $\ddot x = x_2$. Then the equation becomes the first order system $$ \begin{aligned} \dot x_0 &= x_1\,,\\ \dot x_1 &= x_2\,,\\ \dot x_2 &= -x_0 + {x_1}^2 - A\,x_2\,. \end{aligned} $$ This vector field (i.e., ODE) has one singular point, $(x_0,x_1,x_2) = (0,0,0)$, and its linearization at this point has the matrix $$ \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ -1 & 0 & -A \end{pmatrix} $$ The eigenvalues of this matrix are the roots of $\lambda^3-A\lambda^2 - 1 = 0$, and, when $A\ge0$, there is one real root $r>0$ and two complex conjugate roots $z$ and $\bar z$. These roots satisfy $z\bar z r = 1$ and $z\bar z + r(z+\bar z) = 0$, so we have $z+\bar z = -1/r^2 < 0$, so the real part of $z$ is negative. Hence, there is a stable manifold of dimension 2 (and an unstable manifold of dimension 1), say $M^2\subset\mathbb{R}^3$ that passes through $(0,0,0)$ and is such that any initial condition starting on $M$ will spiral in to $(0,0,0)$. Thus, such solutions will stay bounded for all time, in fact, they'll decay to zero.

Actually, no matter how large $A$ is, there will be nonzero solutions that will converge to zero, so you can't prove unboundedness, even though it may be true that the 'generic' solution is unbounded. Here is why: First, convert the system into a first order system in $\mathbb{R}^3$ by setting $x = x_0$, $\dot x = x_1$, and $\ddot x = x_2$. Then the equation becomes the first order system $$ \begin{aligned} \dot x_0 &= x_1\,,\\ \dot x_1 &= x_2\,,\\ \dot x_2 &= -x_0 + {x_1}^2 - A\,x_2\,. \end{aligned} $$ This vector field (i.e., ODE) has one singular point, $(x_0,x_1,x_2) = (0,0,0)$, and its linearization at this point has the matrix $$ \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ -1 & 0 & -A \end{pmatrix} $$ The eigenvalues of this matrix are the roots of $\lambda^3-A\lambda^2 - 1 = 0$, and, when $A\ge0$, there is one real root $r>0$ and two complex conjugate roots $z$ and $\bar z$. These roots satisfy $z\bar z r = 1$ and $z\bar z + r(z+\bar z) = 0$, so we have $z+\bar z = -1/r^2 < 0$, so the real part of $z$ is negative. Hence, there is a stable manifold of dimension 2 (and an unstable manifold of dimension 1), say $M^2\subset\mathbb{R}^3$ that passes through $(0,0,0)$ and is such that any initial condition starting on $M$ will spiral in to $(0,0,0)$. Thus, such solutions will stay bounded for all time, in fact, they'll decay to zero.

Added remark: In fact, even when $A$ is negative, the roots of $\lambda^3-A\lambda^2 - 1 = 0$ are one positive real and either two complex conjugate roots with negative real part or else two real negative roots. Consequently, the stable manifold through $(0,0,0)$ always has dimension $2$, so there are always decaying solutions that are not identically zero.