# 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 $\displaystyle{\frac{\cos A+\cos B - \cos C}{\sin A+\sin B - \sin C}} \geq -\frac{\sqrt{3}}{3}$

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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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thanks I knew there was a reason why they gave $A,B$ and $C$ are less than $120$. –  Kirthi Raman Mar 26 '12 at 12:18

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.

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Notice I have found the minimum of my $f$ with no restriction on $a$ and on $b$, so in principle this is a stronger result than what you wanted (you only wanted to find the minimum of $f$ for pairs $(a,b)$ which come from a triangle). –  Mariano Suárez-Alvarez Mar 26 '12 at 2:44
What if the minimum is at a boundary point, and not in the interior of the regions? –  marty cohen Mar 26 '12 at 3:09
I showed that the unrestricted minimum is $-1/\sqrt{3}$, so the restricted minimum (to whatever restriction you may want to consider) is at least that. This is enough to prove the inequality in the question. –  Mariano Suárez-Alvarez Mar 26 '12 at 3:11