Is the answer 0,0,0 and 2,2,2 only?

Both no

hello, the answer is $(a,b,c)=(0,0,0)$ or $(a,b,c)=(4,4,4)$. Sonnhard.

Hello， you are also wrong

hello, yes only $(a,b,c)=(4,4,4)$, you meant positive integers. Sonnhard.

Dr Sonnhard Graubner wrote: hello, the answer is $(a,b,c)=(0,0,0)$ or $(a,b,c)=(4,4,4)$. Sonnhard. SMOJ wrote: Find, with justification, all positive real numbers $a,b,c$ satisfying the system of equations: \[a\sqrt{b}=a+c,b\sqrt{c}=b+a,c\sqrt{a}=c+b.\]

hello, setting $x=\sqrt{a},y=\sqrt{b},z=\sqrt{c}$ then we get the system $x^2y=x^2+z^2$ $y^2z=y^2+x^2$ $z^2x=z^2+y^2$ eliminating $y,z$ we get for $x$ $x \left( x-2 \right) \left( {x}^{12}-3\,{x}^{11}+6\,{x}^{10}-14\,{x}^ {9}+22\,{x}^{8}-28\,{x}^{7}+37\,{x}^{6}-35\,{x}^{5}+26\,{x}^{4}-21\,{x }^{3}+14\,{x}^{2}-12\,x+8 \right) =0$ with the solutions $x=0$ or $x=2$. Sonnhard.

Clearly, $a(\sqrt{b}-1)=c$ and other cyclic variants. (Note $a, b, c>0$.) Let $a$ be largest among the 3, $a(\sqrt{b}-1)=c$ implies $b \leq 4$. $b(\sqrt{c}-1)=a$ implies $c \geq 4$. Thus, $c \geq b$. $c(\sqrt{a}-1)=b$ implies $a \leq 4$. Thus, $c \geq a$, $c=a=4$. $a(\sqrt{b}-1)=c$ implies $b=4$. Checking, it clearly works. Hence $(a, b, c) = (4, 4, 4)$ is the only solution.