An ellipse bigger than a circle Suppose you have a unit ball in $B^2\subseteq \mathbb{R}^2$ and a point $A=(a,0)$ where $a>\sqrt{2}$.  I would like to show there is an ellipse $E\subseteq\mathrm{conv}(B^2\cup\{A,-A\})$ such that $\mathrm{Vol}(E)>\mathrm{Vol}(B)$. I know it to be true (even in higher dimensions) and I found some proofs using nontrivial analysis. However, I suspect there should be a proof using more elementary math. Can you think of one?
problem illustration http://www.anonimg.com/img/99a813c18ea0b9a2a4f6a9d26a43e2bf.png
If I am correct, the upper right tangential line is $y=-(a^2-1)^{-\frac{1}{2}}(x-a)$ and the point of intersection is $(\frac{1}{a},\frac{\sqrt{a^2-1}}{a})$.
We know that the area of an ellipse is $\mathrm{Vol}(E)=\pi r m$ where $r$ and $m$ is the major and the minor radius of $E$. So we would like to find an ellipse with $rm>1$ as $\mathrm{Vol}(B^2)=\pi$. It would be useful to express $m$ in as a function of $r$, then we could find a point where $f(r)=rm>1$.
The equation of the ellipse would be $\frac{x^2}{r}+\frac{y^2}{m}=1$. It is enought to limit ourselves to the upper right quadrant, where $y=\sqrt{m(1-\frac{x^2}{r})}$. I don't know how to choose $m=g(r)$ so that it has only one intersection with the line $y=-(a^2-1)^{-\frac{1}{2}}(x-a)$.
 A: I think this sketch shows it in a fairly elementary way:
Scale the figure horizontally such that the rhombus made up of the tangent lines become a square. Then the original circle becomes an ellipse with vertical major axis. According to the lemma below, the circle inscribed in the square will have greater area than the ellipse, and it is easy to see that it's within the convex hull. When we scale the figure back, the circle becomes an ellipse that has greater area than the original circle.
Lemma. The largest ellipse that fits within a square is the inscribed circle.
Proof. Clearly the largest ellipse must touch all four sides; if it doesn't we can scale it in the direction where there's room to spare until it does touch one side more, and this makes it larger.
Since the largest ellipse touches all four sides of the square, by symmetry (and a bit of handwaving), its axes must lie on the square's diagonals. Therefore, select a coordinate system such that the square's corners are $(0,\pm1)$ and $(\pm1,0)$; the ellipse will then have the equation $ax^2+by^2=1$ for some $a$ and $b$.
We can find a relation between $a$ and $b$ by noting that the line $y=1-x$ must be tangent to the ellipse, so the discriminant of the equation $ax^2+b(1-x)^2=1$ must be $0$. After some elementary algebra this works out to $a+b=ab$ or $b=\frac{a}{a-1}$.
The area of the ellipse is $\frac{\pi}{4ab}$ so in order to maximize its area we must minimize $ab = \frac{a^2}{a-1}$. A bit of high-school calculus shows that this is minimized for $a=2$, in which case $b=\frac{2}{2-1}=2$, so $a=b$ and the maximal ellipse is indeed a circle.

On the other hand, for the higher-dimensional analogue I don't think it is true that the largest ellipsoid that fits within a right bicone is the inscribed sphere. The relation between $a$ and $b$ in the above calculation would still be the same (carried out in a plane that contains the bicone's axis), but instead of minimizing $ab$ we would want to minimize for example $a^2b=\frac{a^3}{a-1}$, which is not minimized by $a=b=2$.
