Show for all $a,b,c\in\mathbb{N}$, $gcd(a,b)\cdot gcd(b,c) |b\cdot gcd(a,c)$ Really struggling with this one. I've tried using Bezout's Identity to solve this, but algebraic manipulation with the definition of divisibility gets me nowhere. I also considered that $gcd(a,b)|b$ and $gcd(b,c)|b$ implies that $gcd(a,b)gcd(b,c)|b^2$, but that didn't seem to help either since there's no good way to get rid of the square without compromising the proof. Any advice/hints for this one?
 A: Using the Fundamental Theorem of Arithmetic this reduces to prove:
$$\min\{\alpha,\beta\}+\min\{\beta,\gamma\} \le \beta + \min\{\alpha,\gamma\}$$
We may assume, by symmetry, that $\alpha \le \gamma$, so we are left with three cases:

*

*$\beta \le \alpha \le \gamma$

*$\alpha \le \beta \le \gamma$

*$\alpha \le \gamma \le \beta$
And the proof of this cases is straightforward.
A: Proof $1$: $\,\ \begin{align}&\color{#c00}{(a,b)\mid a,b}\\ &\color{#0a0}{(b,c)\:\!\mid b,c}\end{align}\:\!\Rightarrow\: \overbrace{\color{#c00}{(a,b)}\color{#0a0}{(b,c)}}^{\textstyle d}\mid \color{#c00}a\color{#0a0}b,\color{#c00}b\color{#0a0}c\,\Rightarrow\, d\mid(ab,bc)=b(a,c)\ \ $

$\!\!\begin{align} {\rm Proof}\ 2\!:\ (a,b)(b,c) &=(\color{#c00}{ab},\color{#0a0}{bc},\ bb,ac) \mid \color{#c00}{ab},\color{#0a0}{bc}\, \ldots \ \text{ as above.}\\
{\bf or}\  &= (b(a,c),bb,ac)\mid b(a,c)
\end{align}$

Proof $3\!:\  (a,b)(b,c) = (\color{#c00}{ab},bb,ac,\color{#0a0}{bc}) \!\supseteq\! (\color{#c00}{ab},\color{#0a0}{bc}) = b(a,c)\ \ $ [prior using principal ideals]
