[proofplan]
The proof normalizes $K$ by sending its closed John ellipsoid to the closed Euclidean unit ball. John's Ellipsoid Theorem gives sharp inclusions of $K$ between this ellipsoid and a dilation of it, with dilation factor $n$ in general and $\sqrt n$ in the centrally symmetric case. Applying the normalizing affine map converts those closed-ball inclusions into the exact inclusions appearing in the definition of the Banach-Mazur distance.
[/proofplan]
[step:Normalize the John ellipsoid to the Euclidean unit ball]
Let $E \subset K$ be the John ellipsoid of $K$, meaning the unique closed full-dimensional ellipsoid of maximal $n$-dimensional volume contained in $K$. We use John's Ellipsoid Theorem (citing a result not yet in the wiki: John's Ellipsoid Theorem), which applies because $K$ is a convex body and states that if $c \in \mathbb{R}^n$ is the center of $E$, then
\begin{align*}
E \subset K \subset c + n(E-c).
\end{align*}
If $K$ is centrally symmetric, the same theorem gives the sharper inclusion
\begin{align*}
E \subset K \subset c + \sqrt n(E-c).
\end{align*}
Since $E$ is an ellipsoid with non-empty interior, there exists an invertible affine map
\begin{align*}
A: \mathbb{R}^n &\to \mathbb{R}^n
\end{align*}
such that
\begin{align*}
A(E)=\overline{B}(0,1).
\end{align*}
Because $A$ is affine and sends the center $c$ of $E$ to the center $0$ of $\overline{B}(0,1)$, for every $\lambda \geq 0$ one has
\begin{align*}
A\bigl(c+\lambda(E-c)\bigr)=\lambda \overline{B}(0,1).
\end{align*}
[guided]
The geometric normalization is the standard way to pass from John's ellipsoid to a Banach-Mazur estimate. We choose $E \subset K$ to be the John ellipsoid of $K$, that is, the closed full-dimensional ellipsoid of maximal $n$-dimensional volume among all ellipsoids contained in $K$. Let $c \in \mathbb{R}^n$ denote the center of $E$.
We invoke John's Ellipsoid Theorem (citing a result not yet in the wiki: John's Ellipsoid Theorem). Its hypotheses apply because $K$ is a convex body: it is compact, convex, and has non-empty interior. The theorem gives
\begin{align*}
E \subset K \subset c+n(E-c).
\end{align*}
If $K$ is centrally symmetric, John's theorem gives the sharper estimate
\begin{align*}
E \subset K \subset c+\sqrt n(E-c).
\end{align*}
Now we remove the shape of the ellipsoid by an affine change of variables. Since $E$ is an ellipsoid with non-empty interior, there is an invertible affine map
\begin{align*}
A: \mathbb{R}^n &\to \mathbb{R}^n
\end{align*}
with $A(E)=\overline{B}(0,1)$. This map sends the center $c$ of $E$ to the center $0$ of $\overline{B}(0,1)$. Therefore affine linearity gives, for every $\lambda \geq 0$,
\begin{align*}
A\bigl(c+\lambda(E-c)\bigr)=\lambda A(E)=\lambda \overline{B}(0,1).
\end{align*}
This is the point of the normalization: John's inclusion becomes an inclusion between Euclidean balls.
[/guided]
[/step]
[step:Convert the normalized inclusions into the Banach-Mazur bound]
Applying $A$ to the general John inclusion and using preservation of set inclusion under maps gives
\begin{align*}
A(E) \subset A(K) \subset A\bigl(c+n(E-c)\bigr).
\end{align*}
Using $A(E)=\overline{B}(0,1)$ and $A(c+n(E-c))=n\overline{B}(0,1)$, we obtain
\begin{align*}
\overline{B}(0,1) \subset A(K) \subset n\overline{B}(0,1).
\end{align*}
By the defining infimum for $d_{BM}(K,\overline{B}(0,1))$, the admissible affine map $A$ and admissible dilation factor $n$ imply
\begin{align*}
d_{BM}(K,\overline{B}(0,1)) \leq n.
\end{align*}
[guided]
We now apply the affine map $A$ to the inclusions from John's theorem. Since functions preserve inclusions under images, the inclusion
\begin{align*}
E \subset K \subset c+n(E-c)
\end{align*}
implies
\begin{align*}
A(E) \subset A(K) \subset A\bigl(c+n(E-c)\bigr).
\end{align*}
The normalization gives $A(E)=\overline{B}(0,1)$, and affine linearity around the center gives
\begin{align*}
A\bigl(c+n(E-c)\bigr)=n\overline{B}(0,1).
\end{align*}
Therefore
\begin{align*}
\overline{B}(0,1) \subset A(K) \subset n\overline{B}(0,1).
\end{align*}
The definition of $d_{BM}(K,\overline{B}(0,1))$ takes the infimum over all $\lambda \geq 1$ for which some invertible affine map sends $K$ between $\overline{B}(0,1)$ and $\lambda \overline{B}(0,1)$. We have exhibited one such map, namely $A$, with $\lambda=n$. Hence
\begin{align*}
d_{BM}(K,\overline{B}(0,1)) \leq n.
\end{align*}
[/guided]
[/step]
[step:Use the symmetric John inclusion to obtain the sharper factor]
Assume now that $K$ is centrally symmetric. Applying the same invertible affine map $A$ to the symmetric John inclusion gives
\begin{align*}
A(E) \subset A(K) \subset A\bigl(c+\sqrt n(E-c)\bigr).
\end{align*}
Using $A(E)=\overline{B}(0,1)$ and $A(c+\sqrt n(E-c))=\sqrt n\overline{B}(0,1)$, we get
\begin{align*}
\overline{B}(0,1) \subset A(K) \subset \sqrt n\overline{B}(0,1).
\end{align*}
Thus the dilation factor $\sqrt n$ is admissible in the definition of the Banach-Mazur distance, and therefore
\begin{align*}
d_{BM}(K,\overline{B}(0,1)) \leq \sqrt n.
\end{align*}
This proves both claimed estimates.
[/step]
