Maximum value#3

First, we notice that: if $a,b$ are non-negative numbers then

${ \left( a-b \right) }^{ 2 }\left( a+b \right) \ge 0\Longleftrightarrow { a }^{ 3 }+{ b }^{ 3 }\ge { a }^{ 2 }b+a{ b }^{ 2 }\Longleftrightarrow 4\left( { a }^{ 3 }+{ b }^{ 3 } \right) \ge { \left( a+b \right) }^{ 3 }$

$\Longrightarrow a+b\le \sqrt [ 3 ]{ 4\left( { a }^{ 3 }+{ b }^{ 3 } \right) } \quad \left( 1 \right)$

Secondly, in a triangle $ABC$ with 3 angles $A,B,C$ we have

$\sin { A } +\sin { B } =2\cos { \frac { C }{ 2 } } \cos { \frac { A-B }{ 2 } } \le 2\cos { \frac { C }{ 2 } } \quad \left( 2 \right)$

From $(1)$ and $(2)$ ,

$\sqrt [ 3 ]{ \sin { A } } +\sqrt [ 3 ]{ \sin { B } } \le \sqrt [ 3 ]{ 4\left( \sin { A } +\sin { B } \right) } \le \sqrt [ 3 ]{ 4\times 2\cos { \frac { C }{ 2 } } } =2\sqrt [ 3 ]{ \cos { \frac { C }{ 2 } } }$ .

By similarly, $\sqrt [ 3 ]{ \sin { B } } +\sqrt [ 3 ]{ \sin { C } } \le 2\sqrt [ 3 ]{ \cos { \frac { A }{ 2 } } } ,\quad \sqrt [ 3 ]{ \sin { C } } +\sqrt [ 3 ]{ \sin { A } } \le 2\sqrt [ 3 ]{ \cos { \frac { B }{ 2 } } }$

So $\sqrt [ 3 ]{ \sin { A } } +\sqrt [ 3 ]{ \sin { B } } +\sqrt [ 3 ]{ \sin { C } } \le \sqrt [ 3 ]{ \cos { \frac { A }{ 2 } } } +\sqrt [ 3 ]{ \cos { \frac { B }{ 2 } } } +\sqrt [ 3 ]{ \cos { \frac { C }{ 2 } } }$

Or $P\le 1$ . ${ P }_{ min }=1$ when $A=B=C$ or triangle $ABC$ is equilateral.