[proofplan]
We prove the corrected statement with $0\le i<j<n$, since the endpoint $j=n$ would require the undefined exponent $1/(n-j)$ and the suggested interpretation would give a false assertion for large dilates of the unit ball. The proof reduces the comparison to a convexity property of the logarithms of the normalized quermassintegrals. The Alexandrov-Fenchel log-concavity inequalities imply that the sequence $k \mapsto \log(W_k(K)/W_k(B))$ is concave. Since the final normalized quermassintegral equals $1$, concavity forces the endpoint-normalized logarithms $b_k/(n-k)$ to increase with $k$, which gives the desired inequality after exponentiating.
[/proofplan]
[step:Normalize the quermassintegrals by the Euclidean ball]
Let $\mathcal{L}^n$ denote $n$-dimensional Lebesgue measure on $\mathbb{R}^n$, and let $\kappa_n := \mathcal{L}^n(B)$ denote the volume of the Euclidean unit ball. We use the Steiner convention
\begin{align*}
\mathcal{L}^n(K+tB)=\sum_{k=0}^n \binom{n}{k}W_k(K)t^k
\end{align*}
for every $t\geq 0$, where $tB:=\{tx:x\in B\}$ denotes the dilation of $B$ by the scalar $t$. For each $k \in \{0,\dots,n\}$, define the normalized quermassintegral
\begin{align*}
a_k := \frac{W_k(K)}{W_k(B)}.
\end{align*}
Since $B$ is the Euclidean unit ball, its parallel body satisfies $B+tB = (1+t)B$ for every $t\geq 0$, and therefore
\begin{align*}
\mathcal{L}^n(B+tB)
= \kappa_n(1+t)^n
= \sum_{k=0}^n \binom{n}{k}\kappa_n t^k.
\end{align*}
Comparing this with the [Steiner formula](/theorems/4122) gives
\begin{align*}
W_k(B)=\kappa_n
\end{align*}
for every $k \in \{0,\dots,n\}$. Also, by the normalization of quermassintegrals, $W_n(K)=\kappa_n$, so
\begin{align*}
a_n = \frac{W_n(K)}{W_n(B)} = 1.
\end{align*}
[/step]
[step:Use Alexandrov-Fenchel log-concavity for the normalized sequence]
The Alexandrov-Fenchel inequalities for quermassintegrals apply because $K \subset \mathbb{R}^n$ is a convex body, and they give
\begin{align*}
W_k(K)^2 \geq W_{k-1}(K)W_{k+1}(K)
\end{align*}
for every $k \in \{1,\dots,n-1\}$; this is the standard Alexandrov-Fenchel log-concavity inequality for the intrinsic volumes, included here as an external cited input until its Androma theorem entry is added. Since $W_k(B)=\kappa_n$ for every $k$, division by $\kappa_n^2$ gives
\begin{align*}
a_k^2 \geq a_{k-1}a_{k+1}
\end{align*}
for every $k \in \{1,\dots,n-1\}$.
Because $K$ is a convex body, each $W_k(K)>0$, hence each $a_k>0$. Define
\begin{align*}
b_k := \log a_k
\end{align*}
for $k \in \{0,\dots,n\}$. Taking logarithms of the preceding inequality yields
\begin{align*}
2b_k \geq b_{k-1}+b_{k+1}.
\end{align*}
Thus the finite sequence $(b_k)_{k=0}^n$ is concave in the discrete sense.
[guided]
The role of normalization is that the Euclidean ball has the same quermassintegral in every degree: $W_k(B)=\kappa_n$. Therefore the Alexandrov-Fenchel inequality for $W_k(K)$ passes directly to the normalized quantities $a_k=W_k(K)/W_k(B)$.
The inequality
\begin{align*}
W_k(K)^2 \geq W_{k-1}(K)W_{k+1}(K)
\end{align*}
says exactly that the quermassintegrals form a log-concave sequence. Since all $W_k(K)$ are positive for a convex body with nonempty interior, logarithms are legitimate. After defining $b_k:=\log a_k$, the inequality becomes
\begin{align*}
2b_k \geq b_{k-1}+b_{k+1},
\end{align*}
which is the discrete concavity condition. This is the structural input: the desired powers are not obtained by estimating the individual $W_k(K)$ separately, but by reading the slopes of this concave logarithmic sequence.
[/guided]
[/step]
[step:Compare the endpoint slopes of the concave logarithmic sequence]
Since $b_n=\log a_n=0$, the desired corrected inequality is equivalent to
\begin{align*}
\frac{b_i}{n-i} \leq \frac{b_j}{n-j}.
\end{align*}
Indeed, exponentiating this inequality gives the corrected comparison because the exponential function is increasing.
We prove the displayed logarithmic inequality. For a concave finite sequence $(b_k)_{k=0}^n$, the secant slopes are monotone nonincreasing as the left endpoint moves to the right while the right endpoint is fixed. Applying this to the intervals $[i,n]$ and $[j,n]$, with $i<j<n$, gives
\begin{align*}
\frac{b_n-b_i}{n-i} \geq \frac{b_n-b_j}{n-j}.
\end{align*}
Using $b_n=0$, this becomes
\begin{align*}
-\frac{b_i}{n-i} \geq -\frac{b_j}{n-j},
\end{align*}
and hence
\begin{align*}
\frac{b_i}{n-i} \leq \frac{b_j}{n-j}.
\end{align*}
[guided]
Concavity controls slopes. The discrete concavity inequality means that the first differences
\begin{align*}
d_k := b_k-b_{k-1}, \qquad k \in \{1,\dots,n\},
\end{align*}
form a nonincreasing sequence. Indeed,
\begin{align*}
2b_k \geq b_{k-1}+b_{k+1}
\end{align*}
is equivalent to
\begin{align*}
b_k-b_{k-1} \geq b_{k+1}-b_k,
\end{align*}
that is, $d_k \geq d_{k+1}$.
For $m<n$, the secant slope from $m$ to $n$ is the average of the first differences $d_{m+1},\dots,d_n$:
\begin{align*}
\frac{b_n-b_m}{n-m}
=
\frac{1}{n-m}\sum_{\ell=m+1}^n d_\ell.
\end{align*}
Since the sequence $(d_\ell)$ is nonincreasing, the earlier terms $d_{i+1},\dots,d_j$ are greater than or equal to the later tail terms $d_{j+1},\dots,d_n$ in the averaged-order sense. Removing those earlier larger terms can only decrease the remaining tail average. Therefore, for $i<j<n$,
\begin{align*}
\frac{b_n-b_i}{n-i}
\geq
\frac{b_n-b_j}{n-j}.
\end{align*}
Finally $b_n=0$, so this is the same as
\begin{align*}
-\frac{b_i}{n-i}
\geq
-\frac{b_j}{n-j},
\end{align*}
or
\begin{align*}
\frac{b_i}{n-i}
\leq
\frac{b_j}{n-j}.
\end{align*}
This is precisely the logarithmic form of the corrected desired inequality.
[/guided]
[/step]
[step:Exponentiate to recover the quermassintegral comparison]
Because the corrected statement assumes $j<n$, the preceding step gives
\begin{align*}
\frac{\log a_i}{n-i} \leq \frac{\log a_j}{n-j}.
\end{align*}
Exponentiating gives
\begin{align*}
a_i^{1/(n-i)} \leq a_j^{1/(n-j)}.
\end{align*}
Substituting $a_k=W_k(K)/W_k(B)$ yields
\begin{align*}
\left(\frac{W_i(K)}{W_i(B)}\right)^{1/(n-i)}
\leq
\left(\frac{W_j(K)}{W_j(B)}\right)^{1/(n-j)}.
\end{align*}
This is the corrected theorem statement.
[guided]
The logarithmic comparison obtained in the previous step is
\begin{align*}
\frac{\log a_i}{n-i} \leq \frac{\log a_j}{n-j}.
\end{align*}
The exponential map $\exp: \mathbb{R} \to (0,\infty)$ is strictly increasing, so applying it to both sides preserves the inequality. Since $a_i>0$ and $a_j>0$, this gives
\begin{align*}
\exp\left(\frac{\log a_i}{n-i}\right)
\leq
\exp\left(\frac{\log a_j}{n-j}\right),
\end{align*}
that is,
\begin{align*}
a_i^{1/(n-i)} \leq a_j^{1/(n-j)}.
\end{align*}
Finally, the normalized quantities were defined by $a_k=W_k(K)/W_k(B)$, so substitution gives
\begin{align*}
\left(\frac{W_i(K)}{W_i(B)}\right)^{1/(n-i)}
\leq
\left(\frac{W_j(K)}{W_j(B)}\right)^{1/(n-j)}.
\end{align*}
The endpoint $j=n$ is deliberately excluded in the corrected statement: the exponent $1/(n-j)$ is undefined, and interpreting the right-hand side as $1$ would force the false bound $R\leq 1$ for dilates $K=RB$ with $R>1$.
[/guided]
[/step]
