<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] Form an equation with $a,b,c$ as its roots.
We have $ab+bc+ca=-\frac{1}{2}$ from the given relation.
So the equation  with $a,b,c$ as its roots is,
$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=0$
$\Rightarrow x^3-\frac{1}{2}x-abc=0$
If all of $a,b,c$ are of same sign then we have $a=b=c=0$ in which case the inequality is trivially true.So we may consider without loss of generality that $c>0,b<0$ then we have 
$a=-(c+b)\Rightarrow a^2=c^2+b^2+2bc\le c^2+b^2=1-a^2\Rightarrow 2a^2\le 1\Rightarrow \frac{1}{2}-a^2\ge 0$ 
Putting $a$ in the above equation we have $a^3-\frac{1}{2}a+abc=0$
So we have ,
$abc=a(a^2-\frac{1}{2})$
$\Rightarrow a^2b^2c^2=a^2(a^2-\frac{1}{2})^2=a^2(\frac{1}{2}-a^2)^2$
By applying A.M. G.M. on $2a^2(\frac{1}{2}-a^2)^2$(Note that $a^2\ge 0,\frac{1}{2}-a^2\ge 0$ so we can apply A.M.-G.M.) we have,
$2a^2(\frac{1}{2}-a^2)^2\le \left(\frac{2a^2+\frac{1}{2}-a^2+\frac{1}{2}-a^2}{3}\right)^{3}=\frac{1}{27}\Rightarrow a^2b^2c^2=a^2(\frac{1}{2}-a^2)^2\le \frac{1}{27}$
Equality will hold iff $2a^2=\frac{1}{2}-a^2\Rightarrow 3a^2=\frac{1}{2}\Rightarrow a=\frac{1}{\sqrt{6}}$
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