[proofplan] We first reduce by an affine change of variables to the case where the John ellipsoid is the Euclidean unit ball $B(0,1)$. In that normalized position, John's contact decomposition gives contact points $u_i \in \partial K \cap \partial B(0,1)$ and positive weights $c_i$ satisfying both a barycentre identity and an identity decomposition. The supporting inequalities at the contact points constrain every $x \in K$ through the scalars $t_i = u_i \cdot x$. A one-variable quadratic inequality on the interval $[-|x|,1]$, summed against the weights $c_i$, yields $|x|^2 \le n|x|$, hence $|x| \le n$. [/proofplan] [step:Reduce to the case where the John ellipsoid is the unit ball] Write the ellipsoid $E$ in the form \begin{align*} E = a + A B(0,1), \end{align*} where $a \in \mathbb{R}^n$ is its centre and $A: \mathbb{R}^n \to \mathbb{R}^n$ is an invertible [linear map](/page/Linear%20Map). For each scalar $\lambda > 0$, define the dilation of $E$ by factor $\lambda$ about its centre by \begin{align*} \lambda E := a + \lambda A B(0,1). \end{align*} In particular, $nE := a + nA B(0,1)$. Define the affine map \begin{align*} \Phi: \mathbb{R}^n &\to \mathbb{R}^n \\ y &\mapsto A^{-1}(y-a). \end{align*} Then $\Phi(E)=B(0,1)$, and ellipsoids contained in $K$ are carried bijectively by $\Phi$ to ellipsoids contained in $\Phi(K)$. Since affine maps multiply all ellipsoid volumes by the same positive factor $|\det A^{-1}|$, maximality of $E$ in $K$ is equivalent to maximality of $B(0,1)$ in $\Phi(K)$. Thus it is enough to prove that, whenever $B(0,1)$ is the John ellipsoid of a convex body $K \subset \mathbb{R}^n$, one has $K \subset B(0,n)$. Applying this result to $\Phi(K)$ gives \begin{align*} \Phi(K) \subset B(0,n), \end{align*} which is equivalent to \begin{align*} K \subset a + A B(0,n) = a + nA B(0,1) = nE. \end{align*} The inclusion $E \subset K$ is part of the definition of the John ellipsoid. [/step] [step:Use the contact decomposition in John position] Assume from now on that $B(0,1)$ is the John ellipsoid of $K$. We use the contact decomposition part of the [John ellipsoid theorem](/page/John%20Ellipsoid) as an explicit prerequisite in the following form: if a convex body $K \subset \mathbb{R}^n$ has $B(0,1)$ as its John ellipsoid, then there exist an integer $m \in \mathbb{N}$, contact points \begin{align*} u_i \in \partial K \cap \partial B(0,1) \subset \mathbb{R}^n \end{align*} for $1 \le i \le m$, and weights $c_i > 0$ such that \begin{align*} \sum_{i=1}^m c_i u_i &= 0, \\ \sum_{i=1}^m c_i\, u_i \otimes u_i &= I_n. \end{align*} Here $u_i \otimes u_i$ denotes the rank-one linear map $v \mapsto (u_i \cdot v)u_i$, and $I_n$ is the identity map on $\mathbb{R}^n$. Taking traces in the identity decomposition gives \begin{align*} \sum_{i=1}^m c_i \operatorname{tr}(u_i \otimes u_i) = \operatorname{tr}(I_n). \end{align*} Since $|u_i|=1$, we have $\operatorname{tr}(u_i \otimes u_i)=u_i \cdot u_i=1$, while $\operatorname{tr}(I_n)=n$. Therefore \begin{align*} \sum_{i=1}^m c_i = n. \end{align*} Moreover, the same contact theorem provides the supporting inequalities \begin{align*} u_i \cdot y \le 1 \end{align*} for every $y \in K$ and every $1 \le i \le m$. [guided] In John position, the maximal ellipsoid is the unit ball, and the information needed from maximality is encoded by contact points where $K$ touches the unit ball. We use the contact decomposition part of the [John ellipsoid theorem](/page/John%20Ellipsoid) as an explicit prerequisite. Its hypothesis is exactly the normalized hypothesis now in force: $K \subset \mathbb{R}^n$ is a convex body whose John ellipsoid is $B(0,1)$. Under this hypothesis, the theorem supplies an integer $m \in \mathbb{N}$, points \begin{align*} u_i \in \partial K \cap \partial B(0,1) \subset \mathbb{R}^n \end{align*} for $1 \le i \le m$, and positive weights $c_i > 0$ such that \begin{align*} \sum_{i=1}^m c_i u_i &= 0, \\ \sum_{i=1}^m c_i\, u_i \otimes u_i &= I_n. \end{align*} The first identity says that the weighted contact points have barycentre $0$. The second says that the rank-one projections in the contact directions decompose the identity map on $\mathbb{R}^n$. We also need the total mass of the weights. To compute it, take traces in the identity \begin{align*} \sum_{i=1}^m c_i\, u_i \otimes u_i = I_n. \end{align*} The trace is linear, so \begin{align*} \sum_{i=1}^m c_i \operatorname{tr}(u_i \otimes u_i) = \operatorname{tr}(I_n). \end{align*} For each $i$, the contact point satisfies $|u_i|=1$ because $u_i \in \partial B(0,1)$. The rank-one map $u_i \otimes u_i$ has trace $u_i \cdot u_i=|u_i|^2=1$, and $\operatorname{tr}(I_n)=n$. Hence \begin{align*} \sum_{i=1}^m c_i = n. \end{align*} Finally, the contact decomposition is not only an algebraic identity: the contact points come with supporting hyperplanes to $K$. Thus for every $y \in K$ and every $1 \le i \le m$, \begin{align*} u_i \cdot y \le 1. \end{align*} These inequalities are the upper bounds that will control the scalar projections of any point of $K$. [/guided] [/step] [step:Convert the supporting inequalities into a scalar bound for an arbitrary point of $K$] Fix $x \in K$ and define $r := |x|$. For each $1 \le i \le m$, define the scalar \begin{align*} t_i := u_i \cdot x. \end{align*} The supporting inequality gives $t_i \le 1$. Since $|u_i|=1$, the [Cauchy-Schwarz inequality](/theorems/432) in $\mathbb{R}^n$ gives \begin{align*} t_i = u_i \cdot x \ge -|u_i||x| = -r. \end{align*} Thus \begin{align*} t_i \in [-r,1] \end{align*} for every $1 \le i \le m$. For any real number $t \in [-r,1]$, both factors $1-t$ and $t+r$ are nonnegative, so \begin{align*} 0 \le (1-t)(t+r)= -t^2+(1-r)t+r. \end{align*} Equivalently, \begin{align*} t^2 \le (1-r)t+r. \end{align*} Applying this inequality with $t=t_i$ gives \begin{align*} t_i^2 \le (1-r)t_i+r \end{align*} for every $1 \le i \le m$. [guided] We now fix an arbitrary point $x \in K$ and prove that it lies in $B(0,n)$. Define \begin{align*} r := |x|. \end{align*} For each contact direction $u_i$, define the scalar projection \begin{align*} t_i := u_i \cdot x. \end{align*} The supporting hyperplane inequality gives the upper bound \begin{align*} t_i = u_i \cdot x \le 1. \end{align*} For the lower bound, use the Cauchy-Schwarz inequality in the Euclidean inner product on $\mathbb{R}^n$: \begin{align*} u_i \cdot x \ge -|u_i||x|. \end{align*} Since $u_i \in \partial B(0,1)$, we have $|u_i|=1$, and therefore \begin{align*} t_i \ge -r. \end{align*} So each scalar projection satisfies \begin{align*} t_i \in [-r,1]. \end{align*} The useful estimate is now one-dimensional. If $t \in [-r,1]$, then $1-t \ge 0$ and $t+r \ge 0$. Hence their product is nonnegative: \begin{align*} 0 \le (1-t)(t+r). \end{align*} Expanding the product gives \begin{align*} 0 \le -t^2+(1-r)t+r, \end{align*} which is the same as \begin{align*} t^2 \le (1-r)t+r. \end{align*} Applying this with $t=t_i$ is valid because we have just proved $t_i \in [-r,1]$. Therefore, for every $1 \le i \le m$, \begin{align*} t_i^2 \le (1-r)t_i+r. \end{align*} [/guided] [/step] [step:Sum the scalar inequalities using the contact decomposition] Multiply the inequality \begin{align*} t_i^2 \le (1-r)t_i+r \end{align*} by $c_i>0$ and sum over $1 \le i \le m$: \begin{align*} \sum_{i=1}^m c_i t_i^2 \le (1-r)\sum_{i=1}^m c_i t_i + r\sum_{i=1}^m c_i. \end{align*} The left-hand side is \begin{align*} \sum_{i=1}^m c_i t_i^2 &= \sum_{i=1}^m c_i (u_i \cdot x)^2 \\ &= x \cdot \left(\sum_{i=1}^m c_i\, u_i \otimes u_i\right)x \\ &= x \cdot I_n x \\ &= |x|^2 \\ &= r^2. \end{align*} The first sum on the right-hand side is \begin{align*} \sum_{i=1}^m c_i t_i = \sum_{i=1}^m c_i (u_i \cdot x) = \left(\sum_{i=1}^m c_i u_i\right)\cdot x = 0 \cdot x = 0. \end{align*} The second sum is $\sum_{i=1}^m c_i=n$. Hence \begin{align*} r^2 \le rn. \end{align*} [guided] Because all weights satisfy $c_i>0$, multiplying an inequality by $c_i$ preserves its direction. Summing the inequalities \begin{align*} t_i^2 \le (1-r)t_i+r \end{align*} over all contact points gives \begin{align*} \sum_{i=1}^m c_i t_i^2 \le (1-r)\sum_{i=1}^m c_i t_i + r\sum_{i=1}^m c_i. \end{align*} We now evaluate the three sums using the two contact identities. First, since $t_i=u_i \cdot x$, \begin{align*} \sum_{i=1}^m c_i t_i^2 &= \sum_{i=1}^m c_i (u_i \cdot x)^2. \end{align*} The rank-one map $u_i \otimes u_i$ satisfies \begin{align*} x \cdot ((u_i \otimes u_i)x) = (u_i \cdot x)^2. \end{align*} Therefore \begin{align*} \sum_{i=1}^m c_i t_i^2 &= x \cdot \left(\sum_{i=1}^m c_i\, u_i \otimes u_i\right)x \\ &= x \cdot I_n x \\ &= |x|^2 \\ &= r^2. \end{align*} Next, the linear term vanishes because the weighted barycentre of the contact points is zero: \begin{align*} \sum_{i=1}^m c_i t_i &= \sum_{i=1}^m c_i (u_i \cdot x) \\ &= \left(\sum_{i=1}^m c_i u_i\right)\cdot x \\ &= 0. \end{align*} Finally, the trace computation from the previous step gives \begin{align*} \sum_{i=1}^m c_i=n. \end{align*} Substituting these three evaluations into the summed scalar inequality yields \begin{align*} r^2 \le rn. \end{align*} [/guided] [/step] [step:Conclude the containment in the ball of radius $n$] If $r=0$, then $x=0 \in B(0,n)$. If $r>0$, the inequality \begin{align*} r^2 \le rn \end{align*} may be divided by $r$ to obtain \begin{align*} r \le n. \end{align*} Define the closed Euclidean ball of radius $n$ by \begin{align*} \overline{B}(0,n) := \{z \in \mathbb{R}^n : |z| \le n\}. \end{align*} Thus every $x \in K$ satisfies $|x| \le n$, so \begin{align*} K \subset \overline{B}(0,n). \end{align*} This is the normalized containment in John position. By the affine reduction, the general statement is \begin{align*} K \subset nE. \end{align*} Together with $E \subset K$, this proves \begin{align*} E \subset K \subset nE. \end{align*} [/step]