**Note: This is the third version of my answer, one that is, I hope, considerably clearer and cleaner than the previous two.**

This can be reduced to a standard problem in real arithmetic, one that, in principle, is solvable, but just how nice the solution will be is a matter of taste, I think.

For simplicity of notation, let me set $\rho_i = 1/t_i^2$ and $\lambda_i = -c_i/{t_i}^2$ for $i=1,2,3$. Also, set $\rho_0 = (c_1/t_1)^2+(c_2/t_2)^2+(c_3/t_3)^2-1$. Consider the following symmetric matrix depending on a parameter $\nu$
$$
Q(\nu) =
\begin{pmatrix}
\rho_0{+}\nu & \lambda_1 & \lambda_2 &\lambda_3\\
\lambda_1 & \rho_1{-}\nu & 0 & 0\\
\lambda_2 & 0 &\rho_2{-}\nu & 0\\
\lambda_3 & 0 & 0 &\rho_3{-}\nu
\end{pmatrix}
$$
and set $D(\nu) = \det\bigl(Q(\nu)\bigr)$.

*Claims:* (1) The ellipsoid $E$ is disjoint from the unit sphere if and only if there is a value of $\nu$ such that $Q(\nu)$ is either positive or negative definite. (2) The ellipsoid $E$ lies in the interior of the unit ball if and only if the quartic equation $D(\nu)=0$ has $4$ real roots $\nu_i$ that satisfy $0<\nu_0<\nu_1\le\nu_2\le\nu_3$ and, moreover, $Q(\nu)$ is positive definite for some (and, hence, all) $\nu$ satisfying $\nu_0<\nu<\nu_1$.

Note that there are well-established tests for when a matrix is positive definite and for when a quartic polynomial has real roots that are positive. How much you know about the constants $\rho_i$ and $\lambda_i$ will determine how hard actually carrying these tests out will be.

*Here is the argument for the claims:*

Consider the following two quadratic forms on $\mathbb{R}^4$,
$$
q = -{x_0}^2+{x_1}^2+{x_2}^2+{x_3}^2
$$
and
$$
p = \rho_0\,{x_0}^2 + 2\lambda_1\,x_0x_1+2\lambda_2\,x_0x_2+2\lambda_3\,x_0x_3
+ \rho_1\,{x_1}^2+ \rho_2\,{x_2}^2+ \rho_3\,{x_3}^2,
$$

To know whether or not the ellipsoid $E$ and the unit sphere have a real intersection point is the same as knowing whether there is a nonzero vector in $\mathbb{R}^4$ that is a null vector for both $q$ and $p$. Now, it is a standard fact of linear algebra that, on a real vector space of dimension greater than $2$, a pair of quadratic forms has a positive definite linear combination if and only if they have no common null vector other than the zero vector. (See the Footnote for a proof of this standard fact.) This establishes the first claim.

The second claim depends on the first: If $E$ (assumed to have nonempty interior) lies in the interior of the unit ball, then $p$ and $q$ have no common null vector, so, by the first claim, there is some value of $\nu$ for which $Q(\nu)$ is definite (either positive or negative). It then follows from the usual linear algebra proofs that $p$ and $q$ can be simultaneously diagonalized, i.e., (since $q$ and $p$ clearly have type $(3,1)$ and $(1,3)$ respectively), there is a basis of $\mathbb{R}^4$ such that, in the corresponding coordinates $y_i$, we have
$$
q= -{y_0}^2+{y_1}^2+{y_2}^2+{y_3}^2
$$
and
$$
p = -\nu_0\,{y_0}^2+\nu_1\,{y_1}^2+\nu_2\,{y_2}^2+\nu_3\,{y_3}^2
$$
for some (nonzero, since $p$ is nondegenerate) numbers $\nu_i$. Since $E$ is contained in the interior of the unit ball if and only if the closure of the negative cone of $p$ is contained in the negative cone of $q$, it follows that all of the $\nu_i$ must be positive, and, in fact, rearranging $\nu_1,\nu_2,\nu_3$ if necessary, we must have $0<\nu_0<\nu_1\le \nu_2\le \nu_3$. It is now clear that the roots of $D(\nu)=0$ are the $\nu_i$ and that $Q(\nu)$ is positive definite when $\nu_0<\nu<\nu_1$, since $p-\nu q$ is.

Conversely, if $Q(\nu)$ and $D(\nu)$ satisfy the hypotheses, then $p-\nu q$ is a positive definite quadratic form and so the negative cone of $q$ must contain the negative cone of $p$, which implies that $E$ lies in the interior of the unit ball.

**Footnote added on 9/1/13**: To prove the Standard Fact:

One direction is obvious: If the two quadratic forms have a positive definite linear combination, then they have no common null vector.

For the other direction, let $p$ and $q$ be quadratic forms on $\mathbb{R}^n$ (where $n>2$) that have no common zero other than $0\in\mathbb{R}^n$. We need to show that some linear combination of $p$ and $q$ is positive definite.

Consider the map $f = (p,q):\mathbb{R}^n\to\mathbb{R}^2$, which, by hypothesis, sends only the origin to the origin. The normalized map
$$
F(v) = \frac{f(v)}{|f(v)|}
$$
is well-defined and smooth on $\mathbb{R}^n$ minus the origin and is even, i.e., $F(v)=F(-v)$, and homogeneous of degree $0$, so it induces a well-defined smooth map $\phi:\mathbb{RP}^{n-1}\to S^1\subset\mathbb{R}^2$.

Now, the image of $S^{n-1}\subset\mathbb{R}^n$ under $f$ lies in an open halfspace in $\mathbb{R}^2$ if and only if some linear combination of $p$ and $q$ is positive definite, so suppose that this (connected) image does not lie in any open halfspace. Then there will exist two nonzero vectors $x,y\in\mathbb{R}^n$ such that $f(x) = - f(y)\not=(0,0)$. Then, by the usual polarization identity for quadratic forms, one has
$$
f(\cos\theta\,x+\sin\theta\,y)
= \cos2\theta\, f(x) + \sin2\theta\, \tfrac12f(x{+}y)\ \ (\not=0\ \text{for all $\theta$}).
$$
This implies that the path $\gamma(t) = [\cos\theta\,x+\sin\theta\,y]$ for $0\le t\le \pi$, which is a closed path in $\mathbb{RP}^{n-1}$ that generates $H_1(\mathbb{RP}^{n-1},\mathbb{Z})\simeq \mathbb{Z}_2$, is mapped by $\phi$ to a generator of $H_1(S^1,\mathbb{Z})\simeq\mathbb{Z}$, which is absurd.

Note that the proof breaks down for $n=2$ because $H_1(\mathbb{RP}^{1},\mathbb{Z})\simeq \mathbb{Z}$ instead of $\mathbb{Z}_2$. This is good because the pair of quadratic forms $p = x^2-y^2$ and $q = 2xy$ on $\mathbb{R}^2$ have no common null vector and yet have no positive definite linear combination.

arenatural questions in "basic Euclidean geometry" that lead to complicated equations that don't have simple solutions. For example, given $5$ general conics $C_1,\ldots,C_5$ in the plane there's finitely many conics $C$ tangent to all of them, but in general finding $C$ requires solving an equation of degree $3264$ $-$ and even the determination of the degree is nontrivial, never mind the equation! See isites.harvard.edu/fs/docs/icb.topic720403.files/book.pdf $\endgroup$ – Noam D. Elkies Aug 25 '13 at 5:06