<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 Lagrange multiplier proof:
Suppose $a^2b^2c^2$ is maximized. Then since $a + b + c = 0$, two of $a,b$, and $c$ are of one sign, and the third is of the other sign. (Clearly none are zero). Replacing $a,b,$ and $c$ by their negatives if needed, one can assume two are positive, and permuting the variables if necessary one can assume $a,b > 0$ and $c < 0$. 
Then since $c = -(a + b)$, the equation $a^2 + b^2 + c^2 = 1$ becomes
$a^2 + b^2 + (a + b)^2 = 1$ or 
$$a^2 + ab + b^2 = {1 \over 2}$$
And one wants to maximize $a^2b^2c^2$ subject to this contraint for $a, b > 0$. Since $c = -(a + b) < 0$ this is equivalent to maximizing $ab(a + b)$ subject to the constraint, for $a, b > 0$. The Lagrange conditions are 
$$2ab + b^2 = \lambda(2a + b)$$
$$a^2 + 2ab = \lambda(a + 2b)$$
Since $2ab + b^2 = b(2a + b)$, the first equation says $\lambda = b$. Similarly, the second equation implies $\lambda = a$. So $a = b$, and then $a^2 + ab + b^2 = {1 \over 2}$ means $3a^2 = {1 \over 2}$ or $a = {1 \over \sqrt{6}}$. So $a = b = {1 \over \sqrt{6}}$. Then $c = -(a + b) = -{2 \over \sqrt{6}}$, and $a^2b^2c^2 = {4 \over 6^3} = {1 \over 54}$. 
</s>