a, b, c are real numbers such that $$a+b+c=0$$ and $$a^2+b^2+c^2=1$$. Prove that $$a^2b^2c^2 \le \frac{1}{54}$$. When does the equality hold?

From $$(a-b)^2 \ge 0$$, we get $$a^2+b^2+ab \le 3ab$$, and from the constraint $$c=-a-b$$ combined with $$a^2+b^2+c^2=1$$, we derive $$a^2+b^2+ab = \frac{1}{2}$$. Combining these gives $$ab \le \frac{1}{6}$$, and substituting into $$a^2b^2c^2 = a^2b^2\left(\frac{1}{2}-ab\right)$$, treating this as a function of $$ab$$ and maximizing shows the maximum value is $$\frac{1}{54}$$. Equality holds when $$a=b=\frac{1}{\sqrt{6}}$$ and $$c=-\frac{2}{\sqrt{6}}$$.