<s>[INST] $ a+b+c=0,\ a^2+b^2+c^2=1$ implies $ a^4+b^4+c^4=\frac{1}{2}$ This is a high school problem. It is solved by algebraic calculation
(not difficult) - $\ast$.
But I have a problem in interpretation the result.
Here intersection of two surfaces $a+b+c=0,\ a^2+b^2+c^2=1$ is a
circle $C$. That is, by $\ast$, $a^4+b^4+c^4$ is constant on it. What does make
it so ?
Proof : Note that $S:=\{
(a,b,c)|a^4+b^4+c^4=\frac{1}{2}\}$ is a convex surface. In further,
there is $x,\ y\in S$ s.t. $|x|<1<|y|$.
And I guess that $C$ is not a geodesic : Consider a norm on $\mathbb{R}^3$. For instance $\| (a,b,c) \|_1 := |a|+|b|+|c|$ In this case unit ball wrt this norm is octahedron. In further, unit balls of standard Euclidean norm, infinite norm are unit sphere $S_2$, a cube $S_3$, respectively. Clearly, intersection of $S_2$ and plane $\{ (a,b,c)| a+b+c=0\}$ is a geodesic (So is $S_3$). Here considering $S$ and $S_3$, we can assume that $S$ is like $S_3$. That is, $S$ has largest Gaussian curvature at eight points. Hence geodesic is far from these points, but $C$ is not.
I enumerate some these exercises, but how can we interpret the answer ?
[add] Consider $ \Delta':=\{ (x,y,z)\in S_2 | x,\ y,\ z\geq 0\} $, which
is an equilateral geodesic triangle of side length $\frac{\pi}{2}$.
In further $\Delta \subset \Delta'$ is also an equilateral geodesic
triangle of side length $\frac{\pi}{3}$, which touches mid points of
sides in $\Delta'$. Clearly it is not in a plane.
Then we have a claim that if $ f (x,y,z)=(x^4,y^4 ,z^4)$, then
$f(\partial \Delta )$ is in a plane.
[my answer] Consider $SO(3)$-action on $S^2(1)$. Here what is $smallest$ invariant set ? It is $S^2$. Consider a finite subgroup of $SO(3)$. For instance, a group $H$ which acts on cube whose center is origin. So smallest invariant set of $H$ is not $S^2$. It may be union of finite number of great circles, $T$. And note that $F(x,y,z)=x^4+y^4+z^4$ is invariant under $H$. Hence we can guess that $F$ is constant on $T$.  
 [/INST] The geometric approach that you have outlined...  
The intersection of $a+b+c = 0$ and $a^2 + b^2 + c^2 = 1$
A plane cutting a sphere, producing a circle.
$a^4 + b^4 + c^4 = \frac 12$ is a convex surface that resembles a cube with rounded edges and corners.  
As it happens every point on the circle $C$ lies on this "cubeoid."  But, it is still not obvious that this would be the case. 
You could parameterize the circle:
$\begin {bmatrix} 
\frac {\sqrt 3}{3} & \frac {\sqrt 2}{2} &\frac {\sqrt 6}{6}\\
\frac {\sqrt 3}{3} & -\frac {\sqrt 2}{2} &\frac {\sqrt 6}{6}\\
\frac {\sqrt 3}{3} & 0 &-\frac {\sqrt 6}{3}\end{bmatrix}\begin {bmatrix} 0\\\cos\theta\\\sin\theta\end{bmatrix}= \begin {bmatrix} \frac {\sqrt 2}{ 2}\cos\theta + \frac {\sqrt {6}}{6}\sin\theta\\-\frac {\sqrt 2}{ 2}\cos\theta + \frac {\sqrt {6}}{6}\sin\theta\\-\frac {\sqrt 6}{3}\sin\theta\end{bmatrix}$
Would give you a parameterization of the circle.
And $a^4 +b^4 + c^4$ indeed does equal $\frac 12$
I don't think that this is easier than the more algebraic approaches...But it does work.  
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