TITLE: Cosine Inequality
QUESTION [7 upvotes]: 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.

REPLY [7 votes]: Take three vectors $\mathbf a, \mathbf b,$ and $\mathbf c$. We have
$$(\mathbf a + \mathbf b + \mathbf c) \cdot (\mathbf a + \mathbf b + \mathbf c) \ge 0,$$
so
$$\lVert \mathbf a\rVert^2 + \lVert \mathbf b\rVert^2 + \lVert \mathbf c\rVert^2 + 2(\mathbf a \cdot \mathbf b) + 2(\mathbf a \cdot \mathbf c) + 2(\mathbf b \cdot \mathbf c) \ge 0.$$
Now take $a,b,c$ to be the lengths, and $A,B,C$ to be the angles opposite the vectors. Using the fact that $\mathbf a \cdot \mathbf b = \lVert \mathbf a\rVert \cdot \lVert \mathbf b\rVert \cos C$, and similarly for the other two dot products, we get the desired inequality.