A complex equation system

We first notice that these below inequalities hold true with non-negative $(a,b,c)$ :

$4\left( { a }^{ 3 }+{ b }^{ 3 } \right) \ge { \left( a+b \right) }^{ 3 },4\left( { b }^{ 3 }+{ c }^{ 3 } \right) \ge { \left( b+c \right) }^{ 3 },4\left( { c }^{ 3 }+{ a }^{ 3 } \right) \ge { \left( c+a \right) }^{ 3 }$

$\Longrightarrow 4\left( { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 } \right) \ge { \left( a+b \right) }^{ 3 }+{ \left( b+c \right) }^{ 3 }+{ \left( c+a \right) }^{ 3 }$

From the above inequality and AM-GM inequality we then have

${ \left( a+b+c \right) }^{ 3 }={ a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }+3\left( a+b \right) \left( b+c \right) \left( c+a \right) \le { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }+{ \left( a+b \right) }^{ 3 }+{ \left( b+c \right) }^{ 3 }+{ \left( c+a \right) }^{ 3 }\le 9\left( { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 } \right) \left( * \right)$

"=" happens when $a=b=c$ .

Revert to the system of equation

$\begin{cases} \sqrt [ 3 ]{ 9\left( x\sqrt { x } +{ y }^{ 3 }+{ z }^{ 3 } \right) } =x+y+z \quad \left( 1 \right) \\ { x }^{ 2 }+\sqrt { y } =2z \quad \left( 2 \right) \\ \sqrt { y } +{ z }^{ 2 }=\sqrt { 1-x } +2 \quad \left( 3 \right) \end{cases}$

We can easily see that $x,y,z$ satisfy $0\le x\le 1,y\ge 0,z\ge 0$ . Therefore, $x\sqrt { x } ={ \left( \sqrt { x } \right) }^{ 3 }\ge { x }^{ 3 }$

Now, apply the inequality $\left( * \right)$

$9\left( x\sqrt { x } +{ y }^{ 3 }+{ z }^{ 3 } \right) \ge 9\left( { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } \right) \ge { \left( x+y+z \right) }^{ 3 }\Longrightarrow$ LHS $\left( 1 \right)\ge$ RHS $\left( 1 \right)$

$\Longrightarrow \begin{cases} x=y=z \\ \sqrt { x } =x \end{cases}\Longrightarrow x=y=z=0$ or $x=y=z=1$

But $x=y=z=0$ does not satisfy equation $\left( 3 \right)$ .

Hence $\left( x,y,z \right) =\boxed { \left( 1,1,1 \right) }$ is the only real solution of the provided equation system.