<s>[INST] 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
 [/INST] Hint:
$$(a-b+1)^2=a^2+b^2+1-2ab-2b+2a$$



Solution:
$$3a^2+3b^2-2b+2a+1=2a^2+2b^2+(a-b+1)^2+2ab=a^2+b^2+(a+b)^2+(a-b+1)^2>0$$
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