Hello.

Suppose $a$, $b$, $c$ are real numbers such that $a+b+c = 0$ and $a^{2}+b^{2}+c^{2} = 1$.
Prove that $a^{2}b^{2}c^{2}\leq \frac{1}{54}$ and determine the cases of equality.
1. Given the conditions:
   \[
   a + b + c = 0
   \]
   \[
   a^2 + b^2 + c^2 = 1
   \]
   We need to prove that \(a^2 b^2 c^2 \leq \frac{1}{54}\) and determine the cases of equality.

2. From the first condition, we have:
   \[
   c = -(a + b)
   \]
   Substituting \(c\) into the second condition:
   \[
   a^2 + b^2 + (-(a + b))^2 = 1
   \]
   Simplifying:
   \[
   a^2 + b^2 + (a + b)^2 = 1
   \]
   Expanding \((a + b)^2\):
   \[
   a^2 + b^2 + a^2 + 2ab + b^2 = 1
   \]
   \[
   2a^2 + 2b^2 + 2ab = 1
   \]
   Dividing by 2:
   \[
   a^2 + b^2 + ab = \frac{1}{2}
   \]

3. Let \(t = (a + b)^2\). Then:
   \[
   a^2 b^2 c^2 = (ab)^2 (a + b)^2 = (ab)^2 t
   \]
   From the equation \(a^2 + b^2 + ab = \frac{1}{2}\), we can express \(ab\) in terms of \(t\):
   \[
   a^2 + b^2 = t - 2ab
   \]
   Substituting into the equation \(a^2 + b^2 + ab = \frac{1}{2}\):
   \[
   t - 2ab + ab = \frac{1}{2}
   \]
   \[
   t - ab = \frac{1}{2}
   \]
   \[
   ab = t - \frac{1}{2}
   \]

4. Substituting \(ab\) back into \(a^2 b^2 c^2\):
   \[
   a^2 b^2 c^2 = (ab)^2 t = \left(t - \frac{1}{2}\right)^2 t
   \]
   Let \(f(t) = \left(t - \frac{1}{2}\right)^2 t\). We need to find the maximum value of \(f(t)\) for \(t\) in the interval \([0, \frac{2}{3}]\).

5. To find the critical points, we take the derivative of \(f(t)\):
   \[
   f(t) = t \left(t - \frac{1}{2}\right)^2
   \]
   \[
   f'(t) = \left(t - \frac{1}{2}\right)^2 + 2t \left(t - \frac{1}{2}\right)
   \]
   \[
   f'(t) = \left(t - \frac{1}{2}\right) \left(3t - \frac{1}{2}\right)
   \]
   Setting \(f'(t) = 0\):
   \[
   \left(t - \frac{1}{2}\right) \left(3t - \frac{1}{2}\right) = 0
   \]
   \[
   t = \frac{1}{2} \quad \text{or} \quad t = \frac{1}{6}
   \]

6. Evaluating \(f(t)\) at the critical points and endpoints:
   \[
   f\left(\frac{1}{6}\right) = \left(\frac{1}{6} - \frac{1}{2}\right)^2 \cdot \frac{1}{6} = \left(-\frac{1}{3}\right)^2 \cdot \frac{1}{6} = \frac{1}{9} \cdot \frac{1}{6} = \frac{1}{54}
   \]
   \[
   f\left(\frac{2}{3}\right) = \left(\frac{2}{3} - \frac{1}{2}\right)^2 \cdot \frac{2}{3} = \left(\frac{1}{6}\right)^2 \cdot \frac{2}{3} = \frac{1}{36} \cdot \frac{2}{3} = \frac{1}{54}
   \]

7. Therefore, the maximum value of \(f(t)\) is \(\frac{1}{54}\).

8. Equality holds if and only if \(t = \frac{1}{6}\) or \(t = \frac{2}{3}\). This corresponds to the cases:
   \[
   \{a, b, c\} = \left\{\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}}\right\} \quad \text{or} \quad \{a, b, c\} = \left\{-\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}\right\}
   \]

\(\blacksquare\)

The final answer is \( \boxed{ \frac{1}{54} } \)