[proofplan]
We fix the last $n-2$ convex bodies and package the mixed volume as a symmetric [bilinear form](/page/Bilinear%20Form) in the first two arguments. The essential input is Alexandrov's mixed-volume index theorem: for fixed convex bodies $K_3,\dots,K_n$, this bilinear form has Lorentzian signature on the quotient by its kernel, and in particular satisfies the reverse Cauchy--Schwarz inequality on the positive cone of convex bodies. Applying that index inequality to the two convex bodies $K$ and $L$ gives exactly the stated Alexandrov--Fenchel inequality.
[/proofplan]
[step:Fix the background bodies and define the mixed-volume bilinear form]
Let $\mathcal{K}_n$ denote the set of convex bodies in $\mathbb{R}^n$. Fix the convex bodies $K_3,\dots,K_n \in \mathcal{K}_n$ from the statement. Define the symmetric map
\begin{align*}
B: \mathcal{K}_n \times \mathcal{K}_n &\to [0,\infty) \\
(A,C) &\mapsto V(A,C,K_3,\dots,K_n).
\end{align*}
By the symmetry and multilinearity of mixed volume with respect to Minkowski addition and non-negative scalar multiplication, $B$ is symmetric and separately Minkowski-linear on $\mathcal{K}_n$.
[guided]
The goal is to convert the displayed inequality into a two-variable statement. The last $n-2$ convex bodies remain fixed throughout the proof, so we isolate the dependence on the first two entries. Let $\mathcal{K}_n$ be the collection of convex bodies in $\mathbb{R}^n$, and define
\begin{align*}
B: \mathcal{K}_n \times \mathcal{K}_n &\to [0,\infty) \\
(A,C) &\mapsto V(A,C,K_3,\dots,K_n).
\end{align*}
This definition is legitimate for all $A,C \in \mathcal{K}_n$ because $K_3,\dots,K_n$ are convex bodies by hypothesis and mixed volume is defined for $n$ convex bodies in $\mathbb{R}^n$. The symmetry of mixed volume gives $B(A,C)=B(C,A)$, and its multilinearity gives separate Minkowski-linearity in $A$ and $C$. Thus the theorem is exactly the assertion that
\begin{align*}
B(K,L)^2 \ge B(K,K)B(L,L).
\end{align*}
[/guided]
[/step]
[step:Apply Alexandrov's index theorem to the positive cone of convex bodies]
We use Alexandrov's mixed-volume index theorem in the following form: if $K_3,\dots,K_n$ are convex bodies in $\mathbb{R}^n$ and $B(A,C)=V(A,C,K_3,\dots,K_n)$, then for all convex bodies $A,C \in \mathcal{K}_n$,
\begin{align*}
B(A,C)^2 \ge B(A,A)B(C,C).
\end{align*}
The hypotheses of this theorem are satisfied because the fixed bodies $K_3,\dots,K_n$ and the two variable bodies $A,C$ are convex bodies in $\mathbb{R}^n$.
[guided]
The deep geometric input is Alexandrov's mixed-volume index theorem. In the present notation it says: once $K_3,\dots,K_n$ are fixed convex bodies in $\mathbb{R}^n$, the mixed-volume pairing
\begin{align*}
B(A,C)=V(A,C,K_3,\dots,K_n)
\end{align*}
obeys the reverse Cauchy--Schwarz inequality on convex bodies:
\begin{align*}
B(A,C)^2 \ge B(A,A)B(C,C)
\end{align*}
for every pair $A,C \in \mathcal{K}_n$. The theorem applies here because its required data are exactly convex bodies in $\mathbb{R}^n$: the fixed background bodies are $K_3,\dots,K_n$, and the two free inputs are any convex bodies $A$ and $C$. No measure-theoretic integration or limiting argument is being introduced at this stage; those are contained in the proof of Alexandrov's index theorem.
[/guided]
[/step]
[step:Specialize the index inequality to $K$ and $L$]
Taking $A=K$ and $C=L$ in the index inequality gives
\begin{align*}
V(K,L,K_3,\dots,K_n)^2
&= B(K,L)^2 \\
&\ge B(K,K)B(L,L) \\
&= V(K,K,K_3,\dots,K_n)V(L,L,K_3,\dots,K_n).
\end{align*}
This is precisely the desired Alexandrov--Fenchel inequality.
[/step]
