$a≤b≤c$ prime numbers :
$abc\mid{ab+ac+bc-1}\implies a=2$ I would like see explain to this step : 
$$a≤b≤c, \operatorname{  are prime numbers}$$
$1)$ 
$$abc\mid{ab+ac+bc-1}\implies a=2$$ 
$2)$
$$bc\mid{2b+2c-1}\implies b=3$$
My try if correct as following : 
For $1)$
$$ab+bc+ac-1<3bc$$ 
So : 
$$abc\mid{3bc}\implies a\mid 3$$ ?? 
Also for second I have same problem
$$2b+2c-1<4c\implies q\mid{4} ??$$
Can someone explain to me how is this ?? 
 A: I believe this question is related to your earlier question at Prove that : $(ab-1)(ac-1)(bc-1)\equiv 0\pmod{abc}\implies (a,b,c)=(2,3,5)$ and my answer to it. However, you have $a \lt b \lt c$ there instead of $a \le b \le c$, so I assume you want the same thing here as well.
Since you are dealing with positive integers,
$$abc \mid ab+ac+bc-1 \tag{1}\label{eq1A}$$
means that, for some positive integer $k$, you have
$$ab + ac + bc - 1 = k(abc) \tag{2}\label{eq2A}$$
Now, since $k \ge 1$, this means
$$ab + ac + bc - 1 \ge abc \tag{3}\label{eq3A}$$
with the equality occurring only if $k = 1$. Now
$$c \gt a \implies cb \gt ab \tag{4}\label{eq4A}$$
where I multiplied both sides of the inequality by the positive integer $b$, and
$$b \gt a \implies bc \gt ac \tag{5}\label{eq5A}$$
where I multiplied both sides of the inequality with the positive integer $c$. Using these results, along with \eqref{eq3A}, you get
$$\begin{equation}\begin{aligned}
bc + bc + bc - 1 & \gt ab + ac + bc - 1 \\
3bc - 1 & \gt abc \\
3bc - abc & \gt 1 \\
(3-a)bc & \gt 1
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
If $a \ge 3$, then $3 - a \le 0$, so the left hand side of \eqref{eq6A} would be either $0$ or negative. However, as it must be $\gt 1$, this can't be true. As $a$ is a prime, this means $a = 2$ is the only possibility. Also, note using this in \eqref{eq6A} gives $bc \gt 1$, which is true.
As for
$$bc \mid 2b+2c-1 \tag{7}\label{eq7A}$$
you have, as explained earlier for \eqref{eq1A}, that
$$\begin{equation}\begin{aligned}
2b + 2c - 1 & \ge bc \\
2c - 1 & \ge (b - 2)c
\end{aligned}\end{equation}\tag{8}\label{eq8A}$$
If $b \gt 3$, then since $b$ is a prime you have $b \ge 5 \implies b - 2 \ge 3$, but you then get from \eqref{eq8A} that $2c - 1 \ge 3c \implies -1 \ge c$, which is not true. Thus, $b = 3$ is the only possibility available. Note it also works since \eqref{eq8A} then just states that $2c - 1 \ge c \implies c \ge 1$.
Regarding your attempts, for your $1$), I'm not sure where you got
$$ab+bc+ac-1 < 3bc \tag{9}\label{eq9A}$$
As you can see in \eqref{eq3A}, you have that $ab + bc + ac - 1$ is at least the value of $abc$. In \eqref{eq9A}, the inequality is reversed. Also, from this, I'm not sure how you got to
$$abc\mid{3bc}\implies a\mid 3 \tag{10}\label{eq10A}$$
An inequality like in \eqref{eq9A} doesn't imply anything about divisibility, but the reverse can be true, as used for \eqref{eq1A} and \eqref{eq7A}.
Regarding your other statement of
$$2b+2c-1<4c\implies q\mid{4} \tag{11}\label{eq11A}$$
Since $b \lt c$, it's true that $2b + 2c - 1 \lt 4c$. However, I'm not sure what $q$ is and why the inequality would imply that $q \mid 4$.
