1
$\begingroup$

If $A+B+C=\pi$ :

$$ \sin A + \sin B + \sin C \le \frac{3\sqrt{3}}{2} \\ \cos A + \cos B + \cos C \le \frac{3}{2} \\ \tan A + \tan B + \tan C \le 3\sqrt{3} $$

with the equalities holding in the case of an equilateral triangle ($A=B=C=\frac{\pi}{3}$). I've also found out that of all the triangles inscribed in a circle, an equilateral triangle has the largest area.

Why does the maximum of the things I've described exist in the case of an equilateral triangle ? Is it just so or is there a reason for this fact ? Whenever I encounter a question which asks me to maximize something in the case of a triangle, I've taken to simply taking it as an equilateral triangle. Is this safe ? And what are the other situations in which the maximum of something is obtained in the case of an equilateral triangle ?

$\endgroup$
  • 1
    $\begingroup$ I would say that this is not always "safe." Example, maximize the perimeter of a triangle of fixed area 1. This perimeter is unbounded on general triangles, but has a small fixed value for the case of an equilateral triangle. $\endgroup$ – David Mar 27 '17 at 16:19
  • $\begingroup$ Intuitively, I think it is related to the fact that when $x+y+z=C$, for a given $C$, and $x,y,z \geq 0$, then the maximum product of $xyz$ is when $x=y=z$ $\endgroup$ – Χpẘ Mar 27 '17 at 17:45
  • $\begingroup$ @user2460798 So... you are saying that it's true... because equality holds in the AM-GM inequality, only when the numbers are equal. That's an interesting notion. $\endgroup$ – Vishnu V.S Mar 27 '17 at 19:41
0
$\begingroup$

for 1) we have $$\frac{\sin(A)+\sin(B)+\sin(C)}{3}\le \sin\left(\frac{A+B+C}{3}\right)$$ for 2) we have $$\cos(A)+\cos(B)+\cos(C)=1+\frac{r}{R}$$

$\endgroup$
0
$\begingroup$

Because it's cyclic inequality, the answer must be symmetric. Also $\sqrt3$ give a smell of equilateral triangle.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.