[proofplan] The proof is a direct specialization of the Alexandrov-Fenchel inequality for mixed volumes. We choose the two distinguished mixed-volume entries to be $K$ and $B$, and we fill the remaining $n-2$ entries with exactly $n-j-1$ additional copies of $K$ and $j-1$ additional copies of $B$. Symmetry of mixed volume then identifies the three terms in the Alexandrov-Fenchel inequality with $W_j(K)^2$, $W_{j-1}(K)$, and $W_{j+1}(K)$. [/proofplan] [step:Choose the mixed-volume entries that produce the three adjacent quermassintegrals] Fix an integer $j$ with $1 \leq j \leq n-1$. Since $j \leq n-1$, the integer $n-j-1$ is non-negative, and since $j \geq 1$, the integer $j-1$ is non-negative. Let $L_1, \dots, L_{n-2}$ be the following list of convex bodies: $n-j-1$ copies of $K$ followed by $j-1$ copies of $B$. This list has length \begin{align*} (n-j-1) + (j-1) = n-2. \end{align*} Thus the $n$-tuple $(K, B, L_1, \dots, L_{n-2})$ is a valid input for the mixed volume $V$. [guided] Fix an integer $j$ satisfying $1 \leq j \leq n-1$. The Alexandrov-Fenchel inequality compares mixed volumes in which two entries vary and the other $n-2$ entries are held fixed. We want the middle term to contain $n-j$ copies of $K$ and $j$ copies of $B$, because by definition this is $W_j(K)$. To make that happen, we reserve one distinguished slot for $K$ and one distinguished slot for $B$. The remaining slots must then contain $n-j-1$ more copies of $K$ and $j-1$ more copies of $B$. These numbers are non-negative because $j \leq n-1$ and $j \geq 1$. Define $L_1, \dots, L_{n-2}$ to be that list of remaining convex bodies. Its length is \begin{align*} (n-j-1) + (j-1) = n-2, \end{align*} so $(K, B, L_1, \dots, L_{n-2})$ is an $n$-tuple of convex bodies and is therefore a valid mixed-volume input. [/guided] [/step] [step:Apply Alexandrov-Fenchel to the selected entries] The Alexandrov-Fenchel inequality for mixed volumes, applied to the convex bodies $K$, $B$, and $L_1, \dots, L_{n-2}$, gives \begin{align*} V(K, B, L_1, \dots, L_{n-2})^2 \geq V(K, K, L_1, \dots, L_{n-2}) \, V(B, B, L_1, \dots, L_{n-2}). \end{align*} Here the hypotheses are satisfied because $K$, $B$, and each $L_i$ are convex bodies in $\mathbb{R}^n$. This invocation cites a result not yet in the wiki: Alexandrov-Fenchel inequality for mixed volumes. [guided] We now apply the Alexandrov-Fenchel inequality for mixed volumes. In the form needed here, it says that for convex bodies $A$, $C$, and $L_1, \dots, L_{n-2}$ in $\mathbb{R}^n$, \begin{align*} V(A, C, L_1, \dots, L_{n-2})^2 \geq V(A, A, L_1, \dots, L_{n-2}) \, V(C, C, L_1, \dots, L_{n-2}). \end{align*} This cites a result not yet in the wiki: Alexandrov-Fenchel inequality for mixed volumes. We use this theorem with $A := K$ and $C := B$. The required hypotheses are satisfied: $K$ is a convex body by assumption, $B = \overline{B}(0,1)$ is a convex body in $\mathbb{R}^n$, and each $L_i$ is either $K$ or $B$, hence is also a convex body. Therefore \begin{align*} V(K, B, L_1, \dots, L_{n-2})^2 \geq V(K, K, L_1, \dots, L_{n-2}) \, V(B, B, L_1, \dots, L_{n-2}). \end{align*} [/guided] [/step] [step:Identify the three mixed volumes with adjacent quermassintegrals] By symmetry of mixed volume, the first mixed volume contains $1+(n-j-1)=n-j$ copies of $K$ and $1+(j-1)=j$ copies of $B$, so \begin{align*} V(K, B, L_1, \dots, L_{n-2}) = V(K[n-j], B[j]) = W_j(K). \end{align*} The first factor on the right contains $2+(n-j-1)=n-j+1$ copies of $K$ and $j-1$ copies of $B$, hence \begin{align*} V(K, K, L_1, \dots, L_{n-2}) = V(K[n-j+1], B[j-1]) = W_{j-1}(K). \end{align*} The second factor on the right contains $n-j-1$ copies of $K$ and $2+(j-1)=j+1$ copies of $B$, hence \begin{align*} V(B, B, L_1, \dots, L_{n-2}) = V(K[n-j-1], B[j+1]) = W_{j+1}(K). \end{align*} Substituting these three identities into the Alexandrov-Fenchel inequality gives \begin{align*} W_j(K)^2 \geq W_{j-1}(K) W_{j+1}(K). \end{align*} This is the desired log-concavity inequality. [/step]