Let $0<a_1\leq a_2\leq \ldots \leq a_n$ and $n\geq 3$. Define the surface $A$ by the set of points $x=(x_i)$ such that $$\frac{x_1^3}{a_1 }+ \frac{x_2^3}{a_2 }+\ldots+ \frac{x_n^3}{a_n }=1$$ and define the surface $B$ by the points $$x_1^2+ x_2^2+\ldots+ x_n^2=\epsilon,$$ where $\epsilon>0$ is small enough so the surfaces do not intersect. If $$x_n=a_n ( \frac{x_1}{a_1}-\frac{x_2}{a_2 }- \ldots- \frac{x_{n-1}}{a_{n-1} })$$ then $B$ becomes an ellipsoid. In this case, show that the closest point from the surface $A$ to the surface $B$ occurs when $x_n=0$.
Comments
It seems tempting to apply the method of Lagrange multipliers, however in the general case the system of polynomials seems too complicated to provide a full solution.
It is enough to consider nonnegative points, so that surface $A$ becomes $$\frac{|x_1|^3}{a_1 }+ \frac{|x_2|^3}{a_2 }+\ldots+ \frac{|x_n|^3}{a_n }=1.$$ Both surfaces have the same planes of symmetry. That is easy to visualize when $n=3$; the ellipse formed by surface $B$ has a very similar shape to that of surface $A$. How could we use that to show the result?
Example: Surface $A$ with $n=3$, we have $\left( \left| x \right| \right) ^{3}+1/2\, \left( \left| y \right| \right) ^{3}+1/3\, \left( \left| 3\,x-3/2\,y \right| \right) ^{3}=1$. Here $x_3=\left( 3\,x-3/2\,y \right) $. Surface $B$ is given by ${x}^{2}+{y}^{2}+ \left( 3\,x-3/2\,y \right) ^{2}=\sqrt [3]{5}$. We clearly have an ellipse. In this case the ellipse is very close - acctually it touches $A$ in what is supposed to be the closest point. The picture is as follow
This is more a sketch of proof then the full thing.
First, surface B is not just any ellipsoid but the sphere of radius $\sqrt{\varepsilon}$.
Second, if you draw the line that realizes the shortest distance between $A$ and $B$ that that line will be intersect $A$ and $B$ perpendicularly. This is general fact, if it would hit the surfaces in a different angle one can show that it is not a local minimum of the distance between the two surfaces (I think this is essentially a theorem by Almgren, but I can't find a source right now).
Combining these two facts means we can replace surface $B$ by the point $0$ and find the closed distance of $A$ to the origin.
You already noticed that it is sufficient to consider points where $x_i \ge 0$. As the $a_i$ are increasing the point closest to the origin is then the point where $x_1=\sqrt[3]{a_1}$ and $x_2=...=x_n=0$.