Maximum value in triangle

We notice that: $\cos { 2A } +\cos { 2B } +\cos { 2C } =-1-4\cos { A } \cos { B } \cos { C }$

So, $\cos { 2A } +\cos { 2B } +\cos { 2C } \ge -1\quad \Longleftrightarrow \quad \cos { A } \cos { B } \cos { C } \le 0\quad \Longrightarrow \quad \triangle ABC$ is not an acute triangle.

Let $A=max\left\{ A,B,C \right\} \Longrightarrow { 90 }^{ 0 }\le A<{ 180 }^{ 0 }\Longrightarrow \cos { \frac { A }{ 2 } } \le \frac { \sqrt { 2 } }{ 2 } .$

Hence, $\sin { A } +\sin { B } +\sin { C } =\sin { A } +2\cos { \frac { A }{ 2 } } \cos { \frac { B-C }{ 2 } } \le \sin { A } +2\cos { \frac { A }{ 2 } } \le 1+\sqrt { 2 } .$

${ \left( \sin { A } +\sin { B } +\sin { C } \right) }_{ max }=1+\sqrt { 2 }$ when $ABC$ is an isosceles right triangle (for example, $A={ 90 }^{ 0 },B=C={ 45 }^{ 0 }$ ).