<s>[INST] $a+b+c =0$; $a^2+b^2+c^2=1$. Prove that: $a^2 b^2 c^2 \le \frac{1}{54}$ If a,b,c are real numbers satisfying $a+b+c =0; a^2+b^2+c^2=1$. 
Prove that $a^2 b^2 c^2 \le \frac{1}{54}$.
 [/INST] As several have shown above, a, b and c are roots of the cubic $abc=t^3-.5t$. In order for this to have three real roots, abc must lie between the maximum and minimum of $t^3-.5t$. At one of these extrema, two of the roots are equal to the location of the extrema, and the third is found from the requirement that the sum of the roots be zero. One set is $a=b=\frac{1}{\sqrt{6}}$, $c=\frac{-2}{\sqrt{6}}$. Notice that it is not necessary to calculate the discriminant, which can sometimes be messy if the sum of the roots is nonzero.
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