<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] $c=-a-b$, so
$$a^2+b^2+c^2=2a^2+2ab+2b^2=1$$
let $a=x+y,b=x-y$ 
then
$$3x^2+y^2=1/2$$
with the we only consider $ab\le 0$
and $c=-(a+b)=-2x$,
$$c^2=a^2+2ab+b^2\le a^2+b^2=1-c^2$$
so $$8x^2\le 1$$
so $$abc=ab(-a-b)=-(a+b)ab=-2x(x^2-y^2)=-x(8x^2-1)$$
then 
$$a^2b^2c^2=x^2(8x^2-1)(8x^2-1)=\dfrac{1}{16}\cdot 16x^2(1-8x^2)(1-8x^2)\le \dfrac{1}{16}\dfrac{8}{27}=\dfrac{1}{54}$$
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