Factorization Enrichment?

Relevant wiki: Completing The Square

By expanding the equation, we get
$a^2+2b^2+2c^2+4-2ab-2bc-4c=0$ .

After rearranging, the following equation is obtained.
$(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-4c+4)=0$
$(a-b)^2+(b-c)^2+(c-2)^2=0$

Since $a, b, c$ are real numbers, it’s obvious that
$(a-b)^2=(b-c)^2=(c-2)^2=0$
$a=b=c=2$

Therefore, $a+b+c=\boxed{6}$

By the AM - GM inequality,

$\frac {a^2}{2} +b^2 + c^2 +2 = \frac {a^2+b^2}{2} + \frac {b^2+c^2}{2} + \frac {c^2+4}{2} \geq \frac {2|a||b|}{2} + \frac {2|b||c|}{2} + \frac {4|c|}{2} \geq ab+bc+2c$ .

But equality holds, so equality should hold everywhere. Therefore $|a | =|b| = |c| =2$ , and $ab, bc$ and $2c$ are all $\geq 0$ . This happens only when $a=b=c=2$ .