= Solution
Use fixed <masses> in an isolated Newtonian $N$-body system, with origin at its <centre of mass>. Define the $xx$ moment $I_{xx}=\sum_i m_ix_i^2$ and $K_{xx}=\tfrac12\sum_i m_iv_{x,i}^2$. Twice differentiating gives
$$
\frac12\ddot I_{xx}=\sum_i m_iv_{x,i}^2+\sum_i m_ix_i\dot v_{x,i}=2K_{xx}+W_{xx}.
$$
Insert the gravitational <equation of motion> and pair each ordered term $(i,k)$ with $(k,i)$:
$$
\begin{aligned}
W_{xx}&=-G\sum_i\sum_{k\ne i}\frac{m_im_kx_i(x_i-x_k)}{r_{ik}^3}\\
&=-\frac G2\sum_i\sum_{k\ne i}\frac{m_im_k(x_i-x_k)^2}{r_{ik}^3}
=-G\sum_{i<k}\frac{m_im_k(x_i-x_k)^2}{r_{ik}^3}.
\end{aligned}
$$
Thus one component of the <tensor virial theorem> is
$$
\boxed{\frac12\ddot I_{xx}=2K_{xx}-G\sum_{i<k}\frac{m_im_k(x_i-x_k)^2}{r_{ik}^3}.}
$$
A stationary or time-averaged bounded system has $\langle\ddot I_{xx}\rangle=0$, yielding $2\langle K_{xx}\rangle+\langle W_{xx}\rangle=0$. If ordered and random motions are separated, $2K_{xx}=2T_{xx}+\Pi_{xx}$, where $T_{xx}=\tfrac12\int\rho u_x^2dV$ and $\Pi_{xx}=\int\rho\sigma_{xx}^2dV$. Summing the three diagonal equations recovers the scalar <virial theorem>, with the pair <gravitational potential energy> counted once.
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