Distance Between Two Ellipsoids Find the shortest distance between $\begin{align}
&\frac{x_1^2}{a_1^2}+...+\frac{x_n^2}{a_n^2}=1\\
&\frac{x_1^2}{a_1^2}+...+\frac{x_n^2}{a_n^2}=2
\end{align}$.
My thinking is when $n=2$, we have two ellipses where the axes of the second are the same as the first scaled by a factor of $\sqrt{2}$. WLOG, assume $|a_1|<|a_2|$. Then the shortest distance would be $|a_1|(\sqrt{2}-1)$. For the ellipsoid case, I would say the shortest distance is $|a_{\text{min}}|(\sqrt{2}-1)$. I don't know how to prove this formally. I thought about using Lagrange multipliers but had difficulties even setting it up. Any suggestions?

Let $\frac{x_1^2}{a_1^2}+\dots+\frac{x_n^2}{a_n^2} = 1$ and $\frac{y_1^2}{a_1^2}+\dots+\frac{y_n^2}{a_n^2} = 2$. Then by Cauchy-Schwarz
$$ 1 = \sum\limits_{k=1}^{n}{\frac{y_k^2-x_k^2}{a_k^2}} = \sum\limits_{k=1}^{n}{(y_k-x_k)\frac{y_k+x_k}{a_k^2}}\le\left(\sum\limits_{k=1}^{n}{(y_k-x_k)^2}\right)^{\frac{1}{2}}\left(\sum\limits_{k=1}^{n}{\frac{(y_k+x_k)^2}{a_k^4}}\right)^{\frac{1}{2}}. $$
Since $a_k^4\ge(\min{|a_i|})^2a_k^2$ for all $k$, it follows that $\left(\sum\limits_{k=1}^{n}{\frac{(y_k+x_k)^2}{a_k^4}}\right)^{\frac{1}{2}}\le \frac{1}{\min{|a_i|}}\left(\sum\limits_{k=1}^{n}{\frac{(y_k+x_k)^2}{a_k^2}}\right)^{\frac{1}{2}}$, and applying the triangle inequality to the vectors $\vec{x} = \left(\frac{x_1}{a_1},\dots,\frac{x_k}{a_k}\right)$ and $\vec{y} = \left(\frac{y_1}{a_1},\dots,\frac{y_k}{a_k}\right)$ yields
$$ \left(\sum\limits_{k=1}^{n}{\frac{(y_k+x_k)^2}{a_k^2}}\right)^{\frac{1}{2}} = \|\vec{x}+\vec{y}\|\le\|\vec{x}\|+\|\vec{y}\| = 1+\sqrt{2}. $$
Hence, $1\le\frac{1+\sqrt{2}}{\min{|a_i|}}\left(\sum\limits_{k=1}^{n}{(y_k-x_k)^2}\right)^{\frac{1}{2}}$, i.e. $\left(\sum\limits_{k=1}^{n}{(y_k-x_k)^2}\right)^{\frac{1}{2}}\ge(\sqrt{2}-1)\min{|a_i|}$.
If we assume w.l.o.g. that $\min{|a_i|} = |a_1|$, then $(x_1,x_2,\dots,x_n) = (a_1,0,\dots,0)$ and $(y_1,y_2,\dots,y_n) = (\sqrt{2}a_1,0,\dots,0)$ attains equality in the above inequality.
