Let $a>b,a>c$ All $a,b,c$ are odd $a|2c-1<2a \to 2c-1=a \to a=4k-3,c=2k-1,k>1$ $c|2b-1 \to 2b=(2k-1)(2m-1)+1<2a-1=4(2k-1)-3 \to m \leq 2$ $b|2a-1 \to 2km-m-k+1|8k-7$ $m=2: 3k-1|8k-7 \to 3k-1|8(3k-1)-3*(8k-7)=13 \to $ - contradiction $m=1:k|8k-7 \to k=7 \to (a,b,c)=(25,7,13)$

Note that $\left(\dfrac{2a-1}{b}\right)\left(\dfrac{2b-1}{c}\right)\left(\dfrac{2c-1}{a}\right)$ is less than $\dfrac{2a}{b}\dfrac{2b}{c}\dfrac{2c}{a}=8$, and is a product of three odd positive integers. Therefore, at least two of these quotients have to be equal to $1$. Without loss of generality, let $\dfrac{2a-1}{b}=\dfrac{2b-1}{c}=1$. So, $c=4a-3$, then $a|2c-1 \Rightarrow a|8a-7 \Rightarrow a|7$. If $a=1$, we find $(1,1,1)$, and $a=7$ yields $(7,13,25)$, as well as its cyclic permutations $(13,25,7)$ and $(25,7,13)$.