Show that the solution of this (nonlinear) ODE cannot remain bounded as $t\to\infty$

Inspired by the answer of open problem one can say a bit more in general when considering only the $\alpha_i=1$ cases. Although, it is stated that $0 < \alpha_i < 1$, so technically these cases would just barely violate the considered domains for each $\alpha_i$. In these cases the dynamics is linear and can be described with $\dot{x} = M\,x$, with

$$ M = \begin{bmatrix} \lambda_1 + \beta_1 & 1 & 0 & \cdots & 0 \\ \beta_2 & \lambda_2 & 1 & \ddots & \vdots \\ \vdots & 0 & \ddots & \ddots & 0 \\ \beta_{n-1} & \vdots & \ddots & \lambda_{n-1} & 1 \\ \beta_n & 0 & \dots & 0 & \lambda_n \end{bmatrix}. \tag{1} $$

These kind of systems can have non-zero bounded trajectories if $M$ has at least one eigenvalue of zero. One necessary condition for this would be that $\det(M) = 0$, since the determinant of a matrix is equal to the product of its eigenvalues.

It can be shown that in general the determinant of $(1)$ is equal to

$$ \det(M) = \prod_{k=1}^n \lambda_k + \sum_{k=1}^n \left((-1)^{k+1} \beta_k \prod_{m = k+1}^n \lambda_m\right). \tag{2} $$

Even though it holds that $\lambda_i,\beta_i > 0$ for all $i = 1, \cdots, n$, due to the minus signs inside $(2)$ it is possible to have that $\det(M) = 0$ for $n \ge 2$.

For example for $n = 2$ with $\lambda_1,\lambda_2,\beta_1 = 1$ and $\beta_2 = 2$ yields

$$ M = \begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}, \tag{3} $$

which has the eigenvalues $0$ and $3$ and thus can have non-zero bounded trajectories if the initial condition $x(0)$ is chosen such that it doesn't excite the unstable mode associated with the eigenvalue $3$.

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Another sufficient condition for a counter argument would be if a certain system satisfying your description has other equilibria besides the origin. It can be noted that for the linear cases the mode, whose associated eigenvalue is zero, gives a line of equilibria. For all choices for $\alpha_i$ and using $x_1 = 1$ yields $\phi_i(x_1) = \beta_i$. Therefore, a non-zero equilibrium can be constructed by solving $\dot{x} = 0$. In order to split the knowns from the unknowns I define $x' = \begin{bmatrix}x_2 & \cdots & x_n\end{bmatrix}^\top$, such that $\dot{x} = 0$ can be split into $\dot{x}_1 = 0$ and $\dot{x}' = 0$. Substituting $x_1 = 1$ into those two expressions yields

$$ \lambda_1 + x_2 + \beta_1 = 0, \tag{4} $$

$$ A'\,x' + B = 0, \tag{5} $$

with $B = \begin{bmatrix}\beta_2 & \cdots & \beta_n\end{bmatrix}^\top$ and

$$ A' = \begin{bmatrix} \lambda_2 & 1 & 0 & \cdots & 0 \\ 0 & \lambda_3 & 1 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & \lambda_{n-1} & 1 \\ 0 & \cdots & 0 & 0 & \lambda_n \end{bmatrix}. \tag{6} $$

Solving $(5)$ for $x'$ yields $x' = - A'^{-1} B$. Therefore $B$ can be chosen to ensure that $\beta_i > 0$ for $i=2,\cdots,n$. However, this doesn't ensure that $\beta_1 > 0$. Namely, solving $(4)$ for $\beta_1$ yields $\beta_1 = -\lambda_1 - x_2$, where $x_2$ can be obtained from the solution for $x'$. It can be noted that scaling $B$ by a positive scalar $\gamma$ also scales $x'$ by the same scalar. Therefore, if for some valid $B$ one obtains a negative value for $x_2$ one could always find a large enough $\gamma$ such that after scaling $\beta_1$ would become positive. The inverse of $A'$ from $(6)$ can shown to be equal to

$$ A'^{-1}_{ij} = \left\{ \begin{array}{ll} \frac{(-1)^{j-i}}{\prod_{k=i}^j \lambda_k} & \text{if}\ j \geq i \\ 0 & \text{otherwise} \end{array} \right., \tag{7} $$

where $X_{ij}$ denotes the element of matrix $X$ at its $i$th row and $j$th column. Given that each element of $B$ is positive the expression for $x_2$ would thus be a sum of alternating negative and positive terms. Therefore, by choosing some of the odd terms of $B$ sufficiently large would guarantee that the associated solution for $x_2$ would be negative, which thus ensures that $\beta_1$ can be made positive.

For example for $n = 2$ with $\lambda_1,\lambda_2 = 1$ and $\beta_2 = 2$ yields $x_{eq} = \begin{bmatrix}1 & -2\end{bmatrix}^\top$ as equilibrium for every possible $\alpha_i$. It can be noted that due to the fact the the expression for $\dot{x}$ is odd in $x$ also implies that $-x_{eq}$ (thus $\begin{bmatrix}-1 & 2\end{bmatrix}^\top$) would be an equilibrium as well.

However, I am not sure if for $n \geq 2$ these systems always have multiple equilibrium points for any arbitrary choice for $\lambda_i,\beta_i > 0$. But at least I have shown there exists systems that satisfy your description that violate your postulated limits.