Cosine Inequality Show that given three angles $A,B,C\ge0$ with $A+B+C=2\pi$ and any positive numbers $a,b,c$ we have $$bc\cos A + ca \cos B + ab \cos C \ge -\frac {a^2+b^2+c^2}{2}$$
This problem was given in the course notes for a complex analysis course, so I anticipate using  $$bc\cos A + ca \cos B + ab \cos C=\mathbf R\mathbf e (bc\text{cis} A + ca \text{cis} B  + ab \text{cis} C )\qquad=\mathbf R\mathbf e (c\text{cis}(-C)b\text{cis}(-B)+c\text{cis}(-C)a\text{cis}(-A)+a\text{cis}(-A)b\text{cis}(-B))$$
where $\mathbf R\mathbf e(x)$ denotes the real part of $x$, 
is the start, although following this through does not lead me to the inequality I am after.
Also I am unsure of the geometric interpretation of this identity.
This is a similar inequality to  $$bc+ca+ab \le a^2+b^2+c^2$$ 
but the constraints on the angle allow a stronger inequality.
I want some geometric intuition for what the equality is trying to prove, but also I want to see where the inequality comes from.

I applied the Cosine Rule three times (from each side to the opposite angle) and added to arrive at the inequality (actually an equality in a triangle?)
