<s>[INST] Prove that $\displaystyle{\frac{\cos A+\cos B - \cos C}{\sin A+\sin B - \sin C}} \geq -\frac{\sqrt{3}}{3}$ All the angles in a triangle $A,B,$ and $C$ are less than $120^{o}$
Prove that $\displaystyle{\frac{\cos A+\cos B - \cos C}{\sin A+\sin B - \sin C}} \geq -\frac{\sqrt{3}}{3}$
 [/INST] Consider triangle with angles $\small{A_1=120-A,  B_1=120-B, C_1=120-C}$
Applying the triangle inequality in this triangle with angles $A_1, B_1,$ and $C_1$, 
$$\small{B_1 C_1+C_1 A_1 > A_1 B_1}$$
$$\small{\sin A_1 +\sin B_1 > \sin C_1}$$
$$\small{\sin (120-A) +\sin (120-B) > \sin (120-C)}$$
which by applying $\sin(x-y)$ identities
$$\small{\frac{\sqrt{3}}{2}(\cos A+\cos B-\cos C) + \frac{1}{2}(\sin A+ \sin B-\sin C) > 0} \tag{1}$$
And since $\small{a+b > c , \sin A+\sin B - \sin C > 0}$, and therefore dividing by this is perfectly legitimate
Using this observation and re-writing $(1)$, we obtain
$$ \small{\frac{\sqrt{3}}{2} \frac{\cos A + \cos B -\cos C}{\sin A + \sin B - \sin C} + \frac{1}{2} > 0}$$ 
$$ \small{\Rightarrow \frac{\cos A + \cos B -\cos C}{\sin A + \sin B - \sin C}  > \frac{-\sqrt{3}}{3}}$$
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