Note that clearly $a,b,c>1$ and $ab-b+1\mid abc+1-(ab-b+1) = b(ac-a+1)$ implies, together with $(b,ab-b+1)=1$, that $ab-b+1\mid ac-a+1$. Now if $ac-a+1=ab-b+1$ then $a(c-1)=b(a-1)$, so that $a\mid b$, a contradiction. Hence, $ac-a+1\ge 2ab-2b+2$ yielding $ac+2b\ge 2ab+a+1$. Further, $2ab+a+1 = ab+ab+a+1\ge ab+2b+a+1>ab+2b$ that is $ac+2b>ab+2b$. So, $c>b$. Remark. Easy indeed, though took me a while to see.