Points on ellipsoid with maximum Gaussian curvature/mean curvature. 
Find the points on the ellipsoid $$x^2/a^2+y^2/b^2+z^2/c^2=1$$ with maximum Gaussian curvature and mean curvature respectively.

I parametrized it as $(a\sin u\cos v,b\sin u\sin v, c\cos v)$ and managed to compute its Guassian and mean curvature, which are both very messy:$$
K(u,v)=\frac{a^2b^2c^2}{[a^2b^2\cos^2v+c^2(b^2\cos^2u+a^2\sin^2u)\sin^2v]^2 } 
$$
and 
$$H(u,v) =\frac{abc[3(a^2+b^2)+2c^2+(a^2+b^2-2c^2)\cos(2v)-2(a^2-b^2)\cos(2u)\sin^2v]}{8[a^2b^2\cos^2v+c^2(b^2\cos^2u+a^2\sin^2u)\sin^2v]^{3/2}}$$
How can I cleverly find points which maximize the complicated $K$ or $H$? (Perhaps in less than 2 pages of computation, by some geometric reasoning?)  

Using the implicit definition of this surface may be a better approach. Let $$F(x,y,z) = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} - 1.$$ The ellipsoid is then defined by $F(x,y,z) = 0$. Then it is known that the Gaussian curvature at a point $(x,y,z)$ is given by
$$\kappa = \frac{g H^{ad}g^T}{|g|^4}.$$
and the mean curvature is given by
$$h = \frac{gHg^T - \operatorname{trace}(H)|g|^2}{2|g|^3}.$$
where $g = \nabla F$ is the gradient, $H = \nabla g$ is the Hessian matrix, and $H^{ad}$ is its adjoint. Here
$$
g = (2x/a^2, 2y/b^2, 2z/c^2), \, H = \begin{pmatrix} 2/a^2 & 0 & 0 \\ 0 & 2/b^2 & 0 \\ 0 & 0 & 2/c^2 \end{pmatrix} , \quad
H^{ad} = \begin{pmatrix}4/(b^2c^2) & 0 & 0 \\
0 & 4/(a^2c^2) & 0 \\
0 & 0 & 4/(a^2b^2) \end{pmatrix} 
$$
and therefore
$$
|g|^4 = 16\left( (x^2/a^4 + y^2/b^4 + z^2/c^4) \right)^2, \quad
gH^{ad} g^T = 16\frac{x^2/a^2 + y^2/b^2 + z^2/c^2}{a^2b^2c^2} = \frac{16}{a^2b^2c^2}  \, .
$$
The Guassian curvature can now be maximized as follows. We obtain
$$
\kappa = \frac{1}{a^2b^2c^2\left( (x^2/a^4 + y^2/b^4 + z^2/c^4) \right)^2} \, .
$$
Now it is enough to find the extrema of the denominator, subject to the constraint $F(x,y,z) = 0$ which leads to the Lagrange multiplier equations
$$
2x/a^4 = 2\lambda x/a^2, \quad 2y/b^4 = 2\lambda y/b^2, \quad 2z/c^4 = 2\lambda z/c^2\, .
$$
In the generic case where $a,b,c$ are all different, this means $x(\lambda - a^2) = y(\lambda - b^2) = z(\lambda - c^2) = 0$. This is only possible if any two of the unknowns are zero and the third one isn't. Therefore the critical points of $\kappa$ on the ellipsoid are  $(\pm a,0,0), (0,\pm b, 0,0), (0,0,\pm c)$ with Gaussian curvatures $ \frac{a^2}{b^2c^2}, \frac{b^2}{a^2c^2},\frac{c^2}{a^2b^2}$. Pick the largest one and that's the maximum.
