TITLE: How prove this inequality $x_{3}<x_{2}\le\dfrac{3x_{3}}{2}$
QUESTION [5 upvotes]: Let $x_{1},x_{2},x_{3}\in \Bbb{N}\setminus\{0\}$ where
  $$(x_{1},x_{2})+[x_{1},x_{2}]=(x_{1},x_{3})+[x_{1},x_{3}]+1$$
  and such that $\dfrac{x_{i}}{x_{j}}\neq k,k\in N^{+},i,j=1,2,3$.
Show that
  $$x_{3}<x_{2}\le\dfrac{3x_{3}}{2},$$
  where $(x,y)=\gcd(x,y),[x,y]=\mathrm{lcm}(x,y)$.

it is well kown
$$[x,y]=\dfrac{x\cdot y}{(x,y)}$$
so we have
$$x_{1} \left(\dfrac{x_{2}}{(x_{1},x_{2})}-\dfrac{x_{3}}{(x_{1},x_{3})}\right)=(x_{1},x_{3}-x_{2})+1$$
since
$$1\le (x_{1},x_{3})\le \min{(x_{3},x_{1})},1\le(x_{1},x_{2})\le\min{(x_{1},x_{2})}$$
but  only use this can't solve it

REPLY [2 votes]: Let 
$$g = (x_1,x_2) \quad h = (x_1,x_3) \\
x_1 = a_1 g, \quad x_2 = a_2g \quad x_3 = a_3 h
$$
then from $x_i/x_j \not\in \mathbb{N}$ we must have $a_1,a_2,a_3 > 1$ and $x_1/h>1$.
The condition becomes
$$
\begin{align}
g + \frac{x_1 x_2}{g} &= h + \frac{x_1 x_3}{h} + 1 \\
g + a_1 a_2 g =& h + a_1 g a_3 + 1 \\
g(1+a_1(a_2-a_3)) &= h + 1
\end{align}
$$
Case 1: $a_2<a_3$.
This would require $h<0$ so this case is impossible.
Case 2: $a_3<a_2$.
Then
$$
h+1 \ge g(1+a_1) \ge 1+x_1 > 1+h
$$
and this case is also impossible.
Case 3: $a_2=a_3$. Then $g=h+1$. From $g/h = x_2/x_3 \not\in\mathbb{N}$ we must have $h\ge 2$ and
$$
x_3 = (g-1)a_2 < x_2 = \frac{h+1}{h}x_3 \le \frac{3}{2} x_3
$$