$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\leq 1\implies \sum_{cyc}(1+a)(1+b)\leq (1+a)(1+b)(1+c)\implies 2+a+b+c\leq abc$ By AM-GM, $abc\geq 3\sqrt[3]{abc}+2\implies (\sqrt[3]{abc}-2)(\sqrt[3]{abc}+1)^2\geq 0\implies \sqrt[3]{abc}\geq 2\implies abc\geq 8\quad (*)$ \begin{align*} 2\left(\sqrt{\frac{a+b}{ac}}+\sqrt{\frac{b+c}{ba}}+\sqrt{\frac{c+a}{cb}}\right) &=2\left(\frac{\sqrt{(b)(a+b)}+\sqrt{(c)(b+c)}+\sqrt{(a)(c+a)}}{\sqrt{abc}}\right) \\ &\stackrel{\text{C.S.}}{\leq} \frac{2\sqrt{(b+c+a)((a+b)+(b+c)+(c+a))}}{\sqrt{abc}} \\ &= \frac{2\sqrt{2}(a+b+c)}{\sqrt{abc}} \\ &\stackrel{(*)}{\leq} \frac{2\sqrt{2}(a+b+c)}{2\sqrt{2}} \\ &= a+b+c \\ &= \frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-c)(b-a)}+\frac{c^3}{(c-a)(c-b)} \end{align*} Equality will only occur if $abc=8$ so $a+b+c\leq abc-2=6$. By AM-GM, $8=\frac{6^3}{27}\geq\frac{(a+b+c)^3}{27}\geq abc$. Thus, $a=b=c=2$ which is impossible since $a,b,c$ are distinct.