<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] So, $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$
$\implies ab+bc+ca=-\frac12.$
So $a,b,c$ are the roots of the equation $t^3-\frac12t-u=0$ where $u=abc.$
Using discriminant of the cubic equation fro real roots,  $$\triangle= -4\cdot1\cdot\left(-\frac12\right)^3-27(-u)^2=\frac12-27u^2\ge0\iff u^2\le \frac1{54}$$
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