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

$f(a) = 3a^2 +2a + (3b^2-2b+1) \to \triangle' = 1^2 - 3(3b^2-2b+1) = -9b^2 +6b - 2 < 0,
\forall b \text{ since } \triangle' = 3^2 -(-9)(-2) = 9 - 18 = -9 < 0 \to f(a) > 0, \forall a$.
