Is the smallest ellipsoid enclosing a convex set unique? Let $S \subset \mathbb{R}^n$ be a convex set. Assume that it is bounded. We want to find an ellipsoid $E$ of smallest volume such that $S \subset E$. Is $E$ unique?
 A: My claim is that the answer is affirmative by the following 

Lemma. If two different ellipsoids $E_1,E_2$ with the same (hyper-)volume intersect, there is some ellipsoid $E_3$ enclosing $E_1\cap E_2$ with the property that $V(E_3)< V(E_1)$.

Sketch of the proof: I will deal just with the 3D case. $E_1\cap E_2$ is described by something like:
$$ \left \{ \begin{array}{rcl}\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}+\frac{(z-z_0)^2}{c^2}&\leq& 1\\\frac{x^2}{A^2}+\frac{y^2}{B^2}+\frac{z^2}{C^2}&\leq& 1\end{array} \right.\tag{1}$$
where $abc=ABC$. Every point $(x,y,z)$ fulfilling the previous inequalities also fulfills
$$ \frac{(x-x_0)^2}{a^2}+\frac{x^2}{A^2}+\frac{(y-y_0)^2}{b^2}+\frac{y^2}{B^2}+\frac{(z-z_0)^2}{c^2}+\frac{z^2}{C^2}\leq 2, \tag{2} $$
that is the equation of an ellipsoid with volume depending on the product:
$$ \frac{1}{\sqrt{\frac{1}{2a^2}+\frac{1}{2A^2}}}\cdot \frac{1}{\sqrt{\frac{1}{2b^2}+\frac{1}{2B^2}}}\cdot\frac{1}{\sqrt{\frac{1}{2c^2}+\frac{1}{2C^2}}} \tag{3}$$
Through the Cauchy-Schwarz inequality or other means, it is not difficult to show that under the assumption $abc=ABC$, the previous product is $\leq abc$, with equality achieved only at $(a,b,c)=(A,B,C)$, proving the Lemma. Moreover, the approach is dimension-independent.


Back to the original question: assuming that two different ellipsoids $E_1, E_2$ with the same minimal volume enclose $S$, the ellipsoid $E_3$ found through the above "averaging" procedure also encloses $S$, but $V(E_3)<V(E_1)$ contradicts the volume minimality of $E_1$.

This proves a sort of "dual" of John's theorem. I have just learned that the circumscribing ellipsoid with minimal volume is also known as the Loewner-John ellipsoid $E^+$.

The above averaging procedure can also be improved by considering a convex combination of the equations appearing in $(1)$ that leads to the averaged ellipsoid $E_3$ with minimal volume. That probably also leads to an iterative algorithm for finding an enclosing ellipsoid with volume arbitrarily close to the minimum.
