# Why does the maximum of all the trigonometric functions in the case of a triangle always exist in the case of an equilateral triangle?

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 ?

• 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. – David Mar 27 '17 at 16:19
• 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 $x=y=z$ – Χpẘ Mar 27 '17 at 17:45
• @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. – Vishnu V.S Mar 27 '17 at 19:41

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 $$\cos(A)+\cos(B)+\cos(C)=1+\frac{r}{R}$$
Because it's cyclic inequality, the answer must be symmetric. Also $\sqrt3$ give a smell of equilateral triangle.