<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] There are three real numbers, and only two sign(negative, non-negative),So there are two real numbers with same sign,then without loss generality, Assume ab $\ge$ 0  
$1 = a^2+b^2+c^2 = a^2+b^2+(-a-b)^2 = (a^2+b^2+ab) * 2 ≥ 3*ab * 2$
So 
$ab ≤ \dfrac{1}{6}$
$a^2 b^2 c^2 = a^2b^2(a^2+b^2+2ab)=a^2b^2(\dfrac{1}{2}+ab)≤\left(\dfrac{1}{6} \right)^2 *(\dfrac{1}{2}+\dfrac{1}{6})=\dfrac{1}{54}$
Hi, does someone give me any tips about expression format?
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