One day I had a question.
In an acute angled triangle,what is maximum and minimum of $I$?
$$ I=\frac{\sin(A) + \sin(B) + \sin(C)}{\cos(A) + \cos(B) + \cos(C)}$$
Attempt
As law of cosines,
$$\cos(A) + \cos(B) + \cos(C) = \frac{r}{R} + 1$$
$R$ is radius of the circumcircle
$r$ is radius of the incircle
so $ I=\frac{a+b+c}{2(r+R)}$
But I can't evaluate this, could not solve further.
Would you mind solving my question?
Best Answer
Remarks: The maximum of $I$ does not exist for acute triangles. The supremum of $I$ is $2$. The following gives the minimum of $I$. Similarly, we can prove that $I \le 2$.
The minimum of $I$ is $1 + \frac{1}{\sqrt 2}$ when e.g. $A = B = \pi/4$ and $C = \pi/2$.
Proof.
WLOG, assume that $A \le B \le C$. Then $C \ge \pi/3$.
Letting $u = \cos \frac{B - A}{2} \le 1$ and using partial fraction decomposition, we have \begin{align*} I &= \frac{2\cos \frac{C}{2}\,\cos \frac{B - A}{2} + \sin C}{2\sin\frac{C}{2}\, \cos \frac{B - A}{2} + \cos C}\\ &= \frac{2u\cos \frac{C}{2} + \sin C}{2u\sin\frac{C}{2} + \cos C}\\ &= \frac{\cos \frac{C}{2}}{\sin \frac{C}{2}} + \frac{\sin C-\frac{\cos C \, \cos \frac{C}{2}}{\sin \frac{C}{2}}}{2u\sin\frac{C}{2} + \cos C}\\ &\ge \frac{\cos \frac{C}{2}}{\sin \frac{C}{2}} + \frac{\sin C-\frac{\cos C \, \cos \frac{C}{2}}{\sin \frac{C}{2}} }{2\sin\frac{C}{2} + \cos C}\\ & = \frac{2\cos \frac{C}{2} + \sin C}{2\sin\frac{C}{2} + \cos C}\tag{1} \end{align*} where we use $$\sin C -\frac{\cos C \, \cos \frac{C}{2}}{\sin \frac{C}{2}} = \frac{- \cos \frac{3C}{2}}{\sin \frac{C}{2}} \ge 0.$$
Let $f(C) := \frac{2\cos \frac{C}{2} + \sin C}{2\sin\frac{C}{2} + \cos C}$. We have $$f'(C) = - \frac{(1 + \sin \frac{C}{2})(1 - 2\sin\frac{C}{2})^2}{(2\sin\frac{C}{2} + \cos C)^2}\le 0.$$ Thus, we have $$f(C) \ge f(\pi/2) = 1 + \frac{1}{\sqrt 2}. \tag{2}$$
From (1) and (2), we have $I \ge 1 + \frac{1}{\sqrt 2}$.
Also, when $A = B = \pi/4$ and $C=\pi/2$, we have $I = 1 + \frac{1}{\sqrt 2}$.
We are done.