Given the following system of non-linear ODE's: $$ \ddot x_i = \frac{\omega_i^2}{x_i\prod\limits_{j=1}^3 x_j},\quad x_i(0)=1,\quad \dot x_i(0)=0,\quad \text{for $i=1,2,3$}$$ I am interested in the asymptotic behavior. From numerical integration it seems that as $t\rightarrow\infty$ the asymptotic behavior is linear, i.e. $x_i(t)\sim\alpha_it$. Particularly, how do the proportionality constants $\alpha_i$ depend on the parameters $\omega_i$?
In the case where $\omega=\omega_1=\omega_2=\omega_3$ the system has a symmetry, and $x(t)=x_1(t)=x_2(t)=x_3(t)$ satisfies $$ \ddot x = \frac{\omega^2}{x^4},\quad x(0)=1,\quad \dot x(0)=0,$$ which can be solved to find that $x(t)\sim\sqrt{\frac{2}{3}}\omega t$. The question is how this result generalizes to the case were the $\omega_i$ are not equal.
Let us introduce the (scaled) functions $y_i$ with $$x_i = \frac{\omega_i}{(\prod_{j=1}^3 \omega_j )^{1/5}} y_i\;.$$
In terms of these, the equation assume the symmetric form $$\ddot y_i = \frac{1}{y_i \prod_{j=1}^3 y_j} $$ for which your analysis applies.
Note however that the solution with $y_i = y$ that you have found is not the most general solution to the equation and that you need to find an argument why/if the solutions with arbitrary initial conditions will behave symmetrically for asymptotically long times.