We know that all are positive; we also know that all three are greater than 1 because if not, say $c\leq1$. Then $b\sqrt{c}-a\leq b-a<b$, contradiction. We know that if two of the numbers are equal to 4 then the third one is too. So first suppose only one is equal to 4, say $a=4$. Then $c\sqrt{a}-b=2c-b=c$ so $b=c$. Then $a\sqrt{b}-c=4\sqrt{b}-b=4$ so $(b+4)^2=16b$ and $(b-4)^2=0$ so $b=4$, contradiction. So now we assume $a, b, c\neq 4$. By the pigeonhole principle, two of the numbers are either both greater than or both less than 4, say $a, b$. Case 1: $a, b>4$. Then $a=a\sqrt{b}-c>2a-c$ so $c>a>4$. In this way also we can prove $a>b>c>a$, contradiction. Case 2: $a, b<4$. Then $a=a\sqrt{b}-c<2a-c$ so $c<a<4$. In this way also we can prove $a<b<c<a$, contradiction. Therefore the only possible case is $(a, b, c)=(4, 4, 4)$.