For every $x \leq y$ we have by the arithmetic-geometric mean inequality that $\sqrt{y-x} \leq \frac{y-x+1}{2}$ with equality if and only if $y-x=1$. Furthermore, we note that $1-a=b+c$ and the symmetric variants, because $a+b+c=1$. If we apply this to the roots on the left side, we get

$$\begin{aligned} a b \sqrt{b-a}+b c \sqrt{c-b}+a c \sqrt{c-a} & \leq a b \frac{b-a+1}{2}+b c \frac{c-b+1}{2}+a c \frac{c-a+1}{2} \\ & =\frac{1}{2}(a b(2 b+c)+b c(a+2 c)+c a(2 c+b)) \\ & =\frac{1}{4}\left(4 a b^{2}+4 b c^{2}+4 a c^{2}+6 a b c\right) \end{aligned}$$

where in the last step we have already factored out $\frac{1}{4}$ for convenience. Now we note that $a b^{2} \leq b^{3}$ and $a c^{2}+b c^{2} \leq 2 b c^{2}=2 b c \cdot c \leq (b^{2}+c^{2}) c=b^{2} c+c^{3}$. In these inequalities, equality holds respectively if $a=b$ and if $a=b=c$. If we apply this above, we get

$$\begin{aligned} 4 a b^{2}+4 b c^{2}+4 a c^{2}+6 a b c & \leq 3 a b^{2}+b^{3}+3 b c^{2}+3 a c^{2}+b^{2} c+c^{3}+6 a b c \\ & \leq a^{3}+b^{3}+c^{3}+3 a^{2} b+3 a b^{2}+3 b^{2} c+3 b c^{2}+3 a^{2} c+3 a c^{2}+6 a b c \\ & =(a+b+c)^{3}=1 \end{aligned}$$

In the last inequality, equality holds only if the terms $a^{3}, a^{2} b, b^{2} c$ and $a^{2} c$ that we added are all zero. In particular, this implies that $a=0$ and $b^{2} c=0$, from which it follows due to $b \leq c$ that $b=0$.

It is easy to see that not all equality cases can occur simultaneously. For example, for $a=b=0$, the left side of the original inequality is equal to zero.