As the given relations are symmetric in $a,b,c$ , WLOG, we can assume that $a<b<c$ So, $p|bc+1$ $p|ca+1$ $\implies p|c(b-a)$ again, similarly, $p|a(c-b)$ It is clear that $p$ cannot divide $a$ , so $p|c-b$ and $p|b-a$ Now, here's the trick that worked quite well: Therefore I get from the above relations $b-a \ge p \implies b \ge a+p $ Also, $c-b \ge p$ Now, $c \ge p+b \implies c\ge a+2p$ Also, $a \ge 2$, if not then $a=1 \implies p|b+1 ,p|b-1 \implies p|2 $ not possible as $p$ is an odd prime Let us first assume that $a\ge 2$, then we will investigate the case when $a=1$ Plugging $ a\ge 2,b \ge a+p ,c\ge a+2p$ in the given inequality ,we get: $\frac{a+b+c}{3} \ge \frac{3(a+p)}{3} = a+p \ge p+2$ (DONE!!!) Now let me decide about the equality case

Equality holds only when $a=2, b=a+p ,c=a+2p$ Plugging all these into the relations, we have: $p|2p+5 \implies p=5$ and indeed it satisfies!!!! So, the good quadraple for equality will be $(5,2,7,12)$

Oh!! Istek OlympiadTeam edited that no need to think anymore.

what is that solution ohh i am ni$$er