A polynomial inequality in three variables I have a polynomial $P(A,B,C)$ where $A,B,C \in \mathbb{R}$ and $A>0,B>0$. 
$$P(A,B,C)=C^2+A^2-2CA+4AB-2B+C-A$$
When will


*

*$P(A,B,C)>0$

*$P(A,B,C)=0$

*$P(A,B,C)<0$


I have tried factorizing in various ways but none of them is very helpful. Any ideas here?
My attempt:
$P(A,B,C)=(C-A)^2+(C-A)+2B(2A-1)=(C-A)(C-A+1)+2B(2A-1)$
which leads to nowhere

Let $x=C-A$, then $P(A,B,C)=x^2+x+2B(2A-1) \equiv f(x)$.
$$\Delta=1^2-4(2B)(2A-1)=1-8B(2A-1)$$


*

*$\Delta < 0 \implies f(x)>0$,


$$1<8B(2A-1) \implies 2A-1> \frac{1}{8B} \implies A> \frac{8B+1}{16B}$$


*

*$\Delta \ge 0 \land f(x) < 0$,
$$\frac{-1-\sqrt{1-8B(2A-1)}}{2} < C-A < \frac{-1+\sqrt{1-8B(2A-1)}}{2}$$ where $0<A \le \dfrac{8B+1}{16B}$.

*$\Delta \ge 0 \land f(x) = 0$,
$$C-A=\frac{-1 \pm \sqrt{1-8B(2A-1)}}{2}$$  where $0<A \le \dfrac{8B+1}{16B}$.

*$\Delta \ge 0 \land f(x) > 0$,
$$\left| C-A+\frac{1}{2} \right|> \frac{\sqrt{1-8B(2A-1)}}{2}$$ where $0<A \le \dfrac{8B+1}{16B}$.
