<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] Addem, we could do like following. (I write here because it's too long in comment).
1.

$$ a+b+c=0 $$
  $$ (a+b+c)^2=0 $$
  $$ a^2+b^2+c^2+2ab+2bc+2ac=0 $$
  $$ a^2+b^2+c^2=−2(ab+bc+ac) $$
  $$ 1=−2(ab+bc+ac) $$ 
  $$ ⇒ ab+bc+ac=−\frac{1}{2}. $$

2.

$$ (ab+bc+ac)^2=a^2b^2+b^2c^2+a^2c^2+2a^2bc+2ab^2c+2abc^2$$
  $$ (ab+bc+ac)^2=a^2b^2+b^2c^2+a^2c^2+2abc(a+b+c)$$
  $$ (−\frac{1}{2})^2=a^2b^2+b^2c^2+a^2c^2+0$$
  $$ ⇒ a^2b^2+b^2c^2+a^2c^2=\frac{1}{4}. $$

3.

$$ a^2+b^2+c^2=1 $$
  $$ (a^2+b^2 +c^2)^2=1 $$
  $$ a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=1 $$
  $$ a^4+b^4+c^4+2(a^2b^2+b^2c^2+a^2c^2)=1 $$
  $$ a^4+b^4+c^4+2*\frac{1}{4}=1 $$
  $$ ⇒ a^4+b^4+c^4=\frac{1}{2}. $$

</s>