An uncertainty for $a^3+b^3+c^3-3abc$

Question

I have a doubt regarding $a^3+b^3+c^3-3abc$.

For factoring, it is easy that if $a+b+c=0$, then $a^3+b^3+c^3=3abc$ as $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
But on using AM-GM inequality, we see- $$ {a^3+b^3+c^3\over 3}\geq abc \Rightarrow a^3+b^3+c^3\ge3abc $$ AM-GM inequality ensures that equality holds if and only if all variables are equal.
So, equality holds if and only if $a=b=c$, which is trivial.

But we see by factoring that equality holds also if $a+b+c=0$. As the inequality is nothing more than a mere AM-GM, so equality should hold where the inequality ensures us it holds. But it also holds for $a+b+c=0$.
How is it possible and if it is, how can I find such equality cases?

Answer 1

The Arithmetic-Geometric Mean inequality works for only non-negative real numbers. So, for non-negative real numbers equality holds if and only if $a=b=c.$

Answer 2

When $a+b+c=0$, then $a,b,c$ all of $a,b,c$ are not positive, So AM_GM will not be used.

Answer 3

AM-GM only work for non-negative nunber. Where the equality cases hold when all of variables are equal.

On the other hand, if $a,b,c\geqslant 0 $ then $a+b+c \geqslant 0$ if and only if $a=b=c=0$ (if one of them are greater $0$ then $a+b+c>0$).

So the equality also follow by AM-GM.

I hope my answer is useful for you. Thanks.