Solving an algebraic inequality For any $a$, $b$, and $c$ prove
$$3a^2+3b^2-2b+2a+1>0$$
I tried the following
$$(a+1)^2+(b-1)^2+2(a^2+b^2)-1>0\\
(a+1)^2-1+(b-1)^2+2(a^2+b^2)>0\\
(a+1-1)(a+1+1)+(b-1)^2+2(a^2+b^2)>0\\
(a^2+2a)+(b-1)^2+2(a^2+b^2)>0\\
$$
$a^2$ and $(b-1)^2$ and  $2(a^2+b^2)$ are always positive and greater than zero, when $a>0$ our inequality is proved. However when a is negative I can not say $2a$ is always less than or equal to $a^2$. I think this problem should be solved by algebraic identities alone and not by discussing the different signs of $a$? I appreciate your help

$$3a^2+3b^2−2b+2a+1 = (3 a^2 + 2a +\frac{1}{3}) + (3 b^2 - 2b + \frac{1}{3}) + \frac{1}{3} = $$
$$ = (\sqrt{3}a + \frac{1}{\sqrt{3}})^2 + (\sqrt{3}b - \frac{1}{\sqrt{3}})^2 + \frac{1}{3} > 0 $$
