<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] Several good answers have been posted. Just feeling like adding a single variable solution with a geometric twist.
The set of points $(a,b,c)$ is the intersection of the unit sphere and the plane $T:x+y+z=0$ with normal vector $(1,1,1)$. IOW a circle of radius one in the plane $T$ centered at the origin. An occasionally useful factoid is that those points are parametrized by the formula
$$
(a,b,c)=K(\cos t, \cos(t+\frac{2\pi}3), \cos(t-\frac{2\pi}3)),
$$
with the constant $K=\sqrt{2/3}$. Some of you may have seen the relevant trig identities in high school physics in connection with three-phase power: the total power is constant (=> on the unit sphere) and the instantaneous sum of the voltages is zero (=> on the plane $T$).
Anyway, here
$$
bc=K^2\cos(t+\frac{2\pi}3)\cos(t-\frac{2\pi}3)=K^2(\cos^2t-\frac34)
$$
by basic trig identities. Therefore
$$
abc=K^3\frac{4\cos^3t-3\cos t}4=\frac{K^3\cos3t}4.
$$
Hence
$$
(abc)^2\le \frac{K^6}{16}=\frac{8/27}{16}=\frac1{54}.
$$
</s>