= Solution
Write $\langle q\rangle=|\Omega|^{-1}\int_\Omega q\,dx$, $E=\langle\varepsilon\rangle$ and $S=\langle\sigma\rangle$. Use the <infinitesimal strain tensor>, symmetric <stress>, zero <body force>, and the outward unit normal $n$. The <divergence theorem> applied separately to the two displacement derivatives gives
$$
\boxed{E_{ij}=\frac1{2|\Omega|}\int_{\partial\Omega}(u_i n_j+u_j n_i)\,dS.}
$$
For <static elastic equilibrium>, $\sigma_{ik,k}=0$, and therefore $\partial_k(\sigma_{ik}x_j)=\sigma_{ij}$. With <traction> $t_i=\sigma_{ik}n_k$, the <divergence theorem> now gives
$$
\boxed{S_{ij}=\frac1{|\Omega|}\int_{\partial\Omega}t_i x_j\,dS.}
$$
The right-hand side is automatically symmetric because the volume <stress tensor> is symmetric; it can also be written as the integral of $(t_i x_j+t_j x_i)/2$.
To prove the work identity, put $v=u-Ex$. The constant symmetric <tensor> $S$ is equilibrated, so <integration by parts> gives
$$
\begin{aligned}
\int_{\partial\Omega}(\sigma-S)n\cdot v\,dS
&=\int_\Omega(\sigma-S):\operatorname{sym}\nabla v\,dx\\
&=\int_\Omega(\sigma-S):(\varepsilon-E)\,dx\\
&=\int_\Omega\sigma:\varepsilon\,dx-|\Omega|S:E.
\end{aligned}
$$
Here the two cross terms each equal $|\Omega|S:E$, and the last constant term restores one of them. Division by the volume proves the required identity. In particular, affine boundary <displacement> makes $v=0$ on the boundary and yields the <Hill-Mandel energy identity>. The zero-<body force> hypothesis matters: otherwise the first <integration by parts> has an additional volume-force term.
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