[proofplan]
We construct the coefficients by the mixed-volume polynomiality argument inside the convex-body setting. The finite-dimensional input is the polytope case: after a common mixed subdivision, the volume of a Minkowski linear combination of polytopes is a finite sum of determinant expressions and is therefore a homogeneous polynomial of degree n. Then we pass to arbitrary convex bodies by Hausdorff approximation, using continuity of Minkowski addition and volume on bounded families of convex bodies. Finally we distribute each ordinary monomial coefficient among ordered tuples, which gives the required symmetric coefficients.
[/proofplan]
[step:Establish polynomiality for polytopes]
Define
\begin{align*}
\mathcal K^n
:=
\{C\subset \mathbb R^n:C \text{ is nonempty, compact, and convex}\},
\end{align*}
and let $\mathcal L^n$ denote $n$-dimensional Lebesgue measure. First suppose that $P_1,\dots,P_m\in\mathcal K^n$ are polytopes.
We use the Cayley mixed-subdivision theorem for polytopes in the following form. For polytopes $P_1,\dots,P_m\subset\mathbb R^n$, there is a finite set of mixed cells, independent of the positive scalars $t_1,\dots,t_m$, such that for every $t_i>0$ the polytope
\begin{align*}
t_1P_1+\cdots+t_mP_m
\end{align*}
is subdivided into cells of the form
\begin{align*}
t_1Q_1+\cdots+t_mQ_m,
\end{align*}
where each $Q_i$ is a simplex contained in a face of $P_i$ and
\begin{align*}
\dim Q_1+\cdots+\dim Q_m\leq n.
\end{align*}
The cells with dimension sum $n$ have independent direction spaces and account for all $n$-dimensional volume; the remaining cells have zero $n$-dimensional volume. The theorem is applied here exactly under its hypotheses, because $P_1,\dots,P_m$ are polytopes in $\mathbb R^n$.
For one full-dimensional mixed cell, put $d_i:=\dim Q_i$, so $d_1+\cdots+d_m=n$. Choose edge vectors
\begin{align*}
v_{i,1},\dots,v_{i,d_i}
\end{align*}
for the simplex $Q_i$. The standard affine parametrisation of the product of simplices maps onto
\begin{align*}
t_1Q_1+\cdots+t_mQ_m,
\end{align*}
and its derivative columns are
\begin{align*}
t_i v_{i,1},\dots,t_i v_{i,d_i}
\end{align*}
for each $i$. Since the mixed cell is full-dimensional, these $n$ columns span $\mathbb R^n$. Its $n$-dimensional volume is therefore
\begin{align*}
c_Q\,t_1^{d_1}\cdots t_m^{d_m},
\end{align*}
where $c_Q\geq 0$ is the absolute determinant constant divided by the factorials coming from the standard simplex volumes. Summing over the finitely many full-dimensional mixed cells gives a homogeneous polynomial $q_P$ of degree $n$ such that
\begin{align*}
\mathcal L^n(t_1P_1+\cdots+t_mP_m)
=
q_P(t_1,\dots,t_m)
\end{align*}
for $t_i>0$. Both sides are continuous on $[0,\infty)^m$, so the same identity holds for all $t_i\geq 0$.
[/step]
[step:Pass from polytopes to convex bodies]
Let $K_1,\dots,K_m\in\mathcal K^n$. Choose polytopes $P_{i,r}$ with
\begin{align*}
d_H(P_{i,r},K_i)\to 0
\end{align*}
as $r\to\infty$, where $d_H$ is the Hausdorff metric. This is possible because every convex body is the Hausdorff limit of inscribed or circumscribed polytopes. For fixed $R>0$, the Hausdorff convergence is uniform for $(t_1,\dots,t_m)\in[0,R]^m$, because non-negative dilation and Minkowski addition are Lipschitz for $d_H$:
\begin{align*}
d_H(t_1P_{1,r}+\cdots+t_mP_{m,r},t_1K_1+\cdots+t_mK_m)
\leq
R\sum_{i=1}^{m} d_H(P_{i,r},K_i).
\end{align*}
All bodies in these sums lie in one fixed Euclidean ball when $r$ is large and $(t_1,\dots,t_m)\in[0,R]^m$. By the Blaschke selection theorem, the family of convex bodies contained in a fixed Euclidean ball is compact in the Hausdorff metric. Since $\mathcal L^n$ is continuous on convex bodies for Hausdorff convergence, it is uniformly continuous on this compact family. Thus
\begin{align*}
\mathcal L^n(t_1P_{1,r}+\cdots+t_mP_{m,r})
\to
\mathcal L^n(t_1K_1+\cdots+t_mK_m)
\end{align*}
uniformly on $[0,R]^m$.
For each $r$, let $q_r$ be the homogeneous degree-$n$ polynomial associated by the polytope step to the tuple
\begin{align*}
(P_{1,r},\dots,P_{m,r}).
\end{align*}
The vector space of polynomials in $m$ variables of degree at most $n$ is finite-dimensional. Choose a finite unisolvent grid
\begin{align*}
G\subset\{0,1,\dots,n\}^m
\end{align*}
for that space. The coefficients of each $q_r$ are fixed linear combinations of its values on $G$. Uniform convergence on $[0,n]^m$ implies convergence of those grid values, hence convergence of all polynomial coefficients to the coefficients of a polynomial $q$ of degree at most $n$. For any fixed $t\in[0,\infty)^m$, the coefficient convergence gives $q_r(t)\to q(t)$, while Hausdorff convergence gives $q_r(t)\to \mathcal L^n(t_1K_1+\cdots+t_mK_m)$. Therefore
\begin{align*}
\mathcal L^n(t_1K_1+\cdots+t_mK_m)=q(t_1,\dots,t_m)
\end{align*}
for all $t\in[0,\infty)^m$. Homogeneity of $q$ follows from the scaling identity
\begin{align*}
\mathcal L^n(\lambda t_1K_1+\cdots+\lambda t_mK_m)
=
\lambda^n\mathcal L^n(t_1K_1+\cdots+t_mK_m),
\qquad \lambda\geq 0.
\end{align*}
[/step]
[step:Define the ordered coefficients and prove symmetry]
Define the map
\begin{align*}
p:[0,\infty)^m&\to[0,\infty)\\
(t_1,\dots,t_m)&\mapsto \mathcal L^n(t_1K_1+\cdots+t_mK_m).
\end{align*}
By the previous step, $p$ is a homogeneous polynomial of degree $n$. Write its expansion in ordered form as
\begin{align*}
p(t_1,\dots,t_m)
=
\sum_{i_1,\dots,i_n=1}^{m}
V(K_{i_1},\dots,K_{i_n})t_{i_1}\cdots t_{i_n}.
\end{align*}
Let $\alpha=(\alpha_1,\dots,\alpha_m)$ be a multi-index of non-negative integers with
\begin{align*}
\alpha_1+\cdots+\alpha_m=n.
\end{align*}
The ordered expansion is obtained from the ordinary multi-index expansion by dividing the coefficient of
\begin{align*}
t_1^{\alpha_1}\cdots t_m^{\alpha_m}
\end{align*}
equally among the
\begin{align*}
\frac{n!}{\alpha_1!\cdots\alpha_m!}
\end{align*}
ordered tuples containing each index $j$ exactly $\alpha_j$ times. Thus the value assigned to $V(K_{i_1},\dots,K_{i_n})$ depends only on the multiplicities of the entries in the ordered tuple.
If $\sigma$ is a permutation of $\{1,\dots,n\}$, then $(i_{\sigma(1)},\dots,i_{\sigma(n)})$ has the same multiplicities as $(i_1,\dots,i_n)$. Hence
\begin{align*}
V(K_{i_{\sigma(1)}},\dots,K_{i_{\sigma(n)}})
=
V(K_{i_1},\dots,K_{i_n}).
\end{align*}
This proves the required symmetry. Substituting the ordered expansion of $p$ gives exactly
\begin{align*}
\mathcal{L}^n(t_1K_1+\cdots+t_mK_m)
=
\sum_{i_1,\dots,i_n=1}^{m}
V(K_{i_1},\dots,K_{i_n})\,t_{i_1}\cdots t_{i_n},
\end{align*}
with
\begin{align*}
t_iK_i
&:= \{t_i x : x \in K_i\},\\
t_1K_1+\cdots+t_mK_m
&:= \{t_1x_1+\cdots+t_mx_m : x_i \in K_i\}.
\end{align*}
The desired real numbers therefore exist.
[/step]
