TITLE: Conditions for when an off-centre ellipsoid fits inside the unit ball
QUESTION [5 upvotes]: An ellipsoid $E$ has centre $\vec{c}=(c_1,c_2,c_3)$ and semiaxes $t_1$, $t_2$ and $t_3$ aligned with the $x$, $y$ and $z$ axes. What are the necessary and sufficient conditions on $\vec{c}$, $t_1$, $t_2$ and $t_3$ such that $E$ lies inside the unit ball?
For example, we can require that the set of points where the semiaxes meet the ellipsoid surface lie within the unit ball. This gives a set of inequalities such as $(c_1\pm t_1)^2+c_2^2+c_3^2\leq1$, which are necessary but not sufficient. How many points on the surface of the $E$ need to be checked before you know that the whole ellipsoid is within the unit ball? (Clearly, you only need to check the point on the surface of $E$ that is furthest from the origin, but then the problem becomes how to find that point.)
It is easy to formulate a set of inequalities for the special case that the ellipsoid is centred on an axis (i.e. only one $c_i$ is non-zero). However, the off-axis case seems impossible even in 2 dimensions: given an ellipse with centre $\vec c=(c_1,c_2)$ and semiaxes $t_1$ and $t_2$, aligned with the $x$ and $y$ axes, what are the conditions for this to lie within the closed unit disk? 
For the 2D case, the solution might go along these lines (as suggested here amongst other places). Parameterise the perimeter of the ellipse as $x = t_1 \cos\phi + c_1$ and $y = t_2 \sin\phi + c_2$. Then any point on the ellipse perimeter has $x^2+y^2=(t_2^2-t_1^2)\sin^2\phi+2t_1c_1\cos\phi+2t_2 c_2 \sin\phi+\mathrm{const}$. Differentiate this expression to find the $\phi$ for which $x^2+y^2$ is maximal, and see if $x^2+y^2\leq1$ for that $\phi$. If any one of the four parameters $c_1$, $c_2$, $t_1$, $t_2$ is zero then this becomes a simple quadratic, but in general it seems to be not analytically possible.
So, my questions are as follows:

What are the conditions for an off-axis ellipse to lie within the unit disk?
The on-axis ellipse can easily be extended to an on-axis ellipsoid (see Lemma 26 of http://ieeexplore.ieee.org/xpl/articleDetails.jsp?arnumber=669165). Is this also true in the off-axis case?
(I'm not a mathematician so I don't know whether this is a very deep or a completely worthless question...) If the problem is not analytically tractable, why not? How can such an elementary question be so difficult? This question doesn't sound like Fermat's last theorem - it's just very basic Euclidean geometry.

Thanks very much for any help!

REPLY [3 votes]: This is just to elaborate on Robert's idea. Once you bring the question to this form, you can show that the matrix is positive definite iff $1\le \nu\le \min\rho_j$ and
$$
\nu-1\ge\sum_j\frac{\lambda_j^2}{\rho_j-\nu}
$$
(once you fix the first variable, the worst choice for every other one is obvious).
Assume $\rho_1\le\rho_2\le\rho_3$
Note that when $\nu$ is near $1$ and near $\rho_1^-$, the RHS is larger, so the corresponding equation must have two roots (counting with multiplicity) on $(1,\rho_1)$. Since we are also guaranteed a root on each of the intervals $(\rho_j,\rho_{j+1})$ ($j=1,2$), we conclude that a certain quartic monic polynomial in $z=1-\nu$ must have four real negative roots, or, in other words, all roots must be real and all coefficients non-negative. The analytic condition for this event is known. There are several equivalent ways to state it. For programming purposes, the description in this paper seems to be the most convenient (you do not need anything beyond the front page; the definitions, etc. can be googled readily). So, you end up with a lot of elementary algebraic manipulations and 6 inequalities. Whether it beats the purely numeric approach is up to you to decide.