Maximum inscribed sphere inside ellipse and minimum circumscribed sphere containing ellipse Consider the following function
$$
f(x) = \frac{1}{2}x^{\text{T}}Qx + c^{\text{T}}x,
$$
where $Q$ is a real symmetric positive definite $n \times n$ matrix and $c \in \mathbb{R}^{n}$. The ellipse contour of $f$ with level $a \in \mathbb{R}$ can be expressed as
$$
E(a) := \{x \in \mathbb{R}^{n} \mid f(x) = a\}.
$$
The center of $E(a)$ is given by $\hat{x} = -Q^{-1}c$. The function can be now rewritten as
$$
f(x) = \frac{1}{2}(x - \hat{x})^{\text{T}}Q(x - \hat{x}) - \frac{1}{2}c^{\text{T}}Q^{-1}c.
$$
Denote by $S_{\text{ins}}$ the maximum inscribed sphere inside $E(a)$ and $S_{\text{circ}}$ the minimum circumscribed sphere containing $E(a)$. I want to determine the radii $r_{\text{ins}}$ and $r_{\text{circ}}$ of $S_{\text{ins}}$ and $S_{\text{circ}}$, respectively.
Suppose the eigenvalues of $Q$ are ranked in an ascending order, i.e.,
$$
0 < \lambda_1 \leq \lambda_2 \leq \dots \leq \lambda_n.
$$
In the paper, they said the radius are given by
$$
r_{\text{ins}} = \sqrt{\frac{2(a-t)}{\lambda_n}}
$$
and
$$
r_{\text{circ}} = \sqrt{\frac{2(a-t)}{\lambda_1}},
$$
where $t = - \frac{1}{2}c^{\text{T}}Q^{-1}c$. But they give no proof. Can someone please explain why this is true? Here is the link of the paper:
https://link.springer.com/article/10.1007/s10898-011-9713-2
 A: Change coordinates by defining $y = x - \hat{x}$. Now your function is
$$
g(y) = \frac12 y^t Q y + t,
$$
where $t = -\frac12 c^t Q^{-1} c$.
The level set for $g(y) = a$ is then all points $y$ with
$$
y^t Q y = 2(a - t)
$$
Because $Q$ is symmetric positive definite matrix, there's an orthogonal matrix $R$ whose rows are the (unit) eigenvectors of $Q$, such that
$$
Q = R^t D R
$$
where $D = diag(\lambda_1, \ldots, \lambda_n)$. So we can rewrite $g$ as
$$
g(y) = y^t R^t D R y + t.
$$
Once again changing coordinates to $z = Ry$, we have
$$
h(z) = z^t D z + t
$$
whose level-set, for $a$, is
$$
\{z \mid z^t D z  = 2(a-t) \}
$$
Writing that out, we have
$$
z_1^2 \lambda_1 + \ldots + z_n^2 \lambda_n = 2(a-t)
$$
Now because of the ordering of the $\lambda_i$, we can say
$$
z_1^2 \lambda_1 + \ldots + z_n^2 \lambda_n \ge 
z_1^2 \lambda_1 + \ldots + z_n^2 \lambda_1 = \lambda_1 (z_1^2 + z_n^2) \tag{1}
$$
so
$$
\lambda_1 \|z\|^2 \ge 2(a-t) 
$$
hence
$$
\|z\|^2 \ge \frac{2(a-t)}{\lambda_1 } 
$$
so
$$
|z| \ge \sqrt{\frac{2(a-t)}{\lambda_1 }}.
$$
which says that every point on the ellipsoid is at least that far from the origin (with $(1,0,\ldots, 0)$ being exactly that far from the origin), hence the radius of the inscribed sphere must be that number.
I'll bet that you can take equation 1 and write a less-than-or-equal version involving $\lambda_n$, and derive the other half of the result for yourself.
A: If $u=x-\hat x$, then we must find maximum and minimum of the function $\sqrt{u^Tu}$, subject to the constraint
$${1\over2}u^TQu=a-t.$$
If $\alpha$ is a Lagrange multiplier, we must then find the stationary points of
$$
F(u)=u^Tu+{1\over2}\alpha u^TQu,
$$
i.e. the values of $u$ which make the gradient of $F$ vanish:
$$
{\partial F\over \partial u}=2u+\alpha Qu=0,
$$
which is the same as
$$
Qu=-{2\over\alpha}u.
$$
Hence the stationary points are eigenvectors $u_i$ of $Q$ and $\alpha=-2/\lambda_i$.
The norm of $u_i$ can be found from the constraint equation:
inserting there $u=u_i$ we obtain
$${1\over2}u_i^TQu_i=a-t,
\quad\text{that is:}\quad
u_i^Tu_i={2(a-t)\over\lambda_i}.
$$
Maximum and minimum of $\sqrt{u^Tu}$ are then
$$
\sqrt{2(a-t)\over\lambda_\min}\quad\text{and}\quad\sqrt{2(a-t)\over\lambda_\max}.
$$
