All the angles in a triangle $A,B,$ and $C$ are less than $120^{o}$
Prove that $\displaystyle{\frac{\cos A+\cos B - \cos C}{\sin A+\sin B - \sin C}} \geq -\frac{\sqrt{3}}{3}$
All the angles in a triangle $A,B,$ and $C$ are less than $120^{o}$
Prove that $\displaystyle{\frac{\cos A+\cos B - \cos C}{\sin A+\sin B - \sin C}} \geq -\frac{\sqrt{3}}{3}$
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}}$
Some brute force...
Let $f$ be your quotient, depending on $a$ and $b$ (I have eliminated $c$, because $c=\pi-a-b$):
In[1]:= f = (Cos[a] + Cos[b] - Cos[Pi - a - b]) / (Sin[a] + Sin[b] - Sin[Pi - a - b]) Cos[a] + Cos[b] + Cos[a + b] Out[1]= ---------------------------- Sin[a] + Sin[b] - Sin[a + b]
Notice that Mma is being smart and has midly rewritten our expression...
Find the points where the gradient of $f$ vanishes:
In[2]:= Reduce[{D[f, a] == 0, D[f, b] == 0}, {a, b}] Out[2]= (C[1] | C[2]) \[Element] Integers && -2 Pi -2 Pi > ((a == ----- + 2 Pi C[1] && b == ----- + 2 Pi C[2]) || 3 3 2 Pi 2 Pi > (a == ---- + 2 Pi C[1] && b == ---- + 2 Pi C[2])) 3 3
I used Reduce
and not Solve
, because the latter complains a little, harmlessly in this case; Reduce
gives the answer in the annoying form above, though. In any case, this tells us that the critical points of the function occur at the points of the form $(-2\pi/3+2\pi n,-2\pi/3+2\pi m)$ and $(2\pi/3+2\pi n,2\pi/3+2\pi m)$ wit $n$, $m\in\mathbb Z$. To find the value of $f$ at these points it sufficies to take $n=m=0$, because $f$ is periodic in $a$ and in $b$.
In[3]:= f /. {a -> -2 Pi/3, b -> -2 Pi/3} 1 Out[3]= ------- Sqrt[3] In[4]:= f /. {a -> 2 Pi/3, b -> 2 Pi/3} 1 Out[4]= -(-------) Sqrt[3]
This tells us that the minimal value of $f$ is $-1/\sqrt{3}$ and proves what you want.