<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] $$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$
$$\implies 0^2=1+2 (ab+bc+ca)\implies ab+bc+ca=-\frac12$$
Again, $ab+bc+ca=a(b+c)+bc=a(-a)+bc\implies bc-a^2$
$$\implies bc-a^2=-\frac12\implies bc=\frac{2a^2-1}2$$
$$\implies abc=a\cdot \frac{2a^2-1}2=\frac{2a^3-a}2$$
Now use  Second derivative test for the extreme values of $f(a)=\frac{2a^3-a}2$ 
As $a^2=1-(b^2+c^2)\le1, -1\le a\le 1$

Alternatively,
we can set $a=\cos A, b=\sin A\cos B,c=\sin A\sin B$
As $a+b+c=0,\cos A+\sin A\cos B+\sin A\sin B=0\implies \cos A=-\sin A(\cos B+\sin B) $
Squaring  we get, $\cot^2A=1+2\sin B\cos B$
$$abc=\cos A\cdot\sin A\cos B\cdot\sin A\sin B=\cos A\sin^2A(\sin B\cos B)$$
$$=\cos A\sin^2A\cdot \frac{\cot^2A-1}2=\cos A\cdot\frac{\cos^2A-\sin^2A}2=\frac{2\cos^3A-\cos A}2 $$ which can be handled like the previous method
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