Is it possible to proof: when $gcd\left(m_1,m_2,m_3\right)=1$, then $\frac{lcm\left(m_1,m_2\right)m_3}{gcd\left(lcm\left(m_1,m_2\right),m_3\right)}=\frac{lcm\left(m_1,m_2,m_3\right)}{gcd\left(m_1,m_2,m_3\right)}$ is true for any positive integers $ m_1,m_2, m_3 $?
I have test many numbers looks it is actually true.
Your conjecture is indeed correct. I suppose you can prove the following:
$$1) \ \ ab = gcd(a,b)\cdot lcm(a,b)$$ $$2) \ \ lcm(a,b,c) = lcm(lcm(a,b), c)$$ These are standard results, you can look them up.
From (1) you get:
$$gcd(lcm(a,b),c) \cdot lcm(lcm(a,b),c) = lcm(a,b)\cdot c$$ or that $$gcd(lcm(a,b),c) = \frac{lcm(a,b)\cdot c}{lcm(a,b,c)}$$ Now consider: $$\frac{lcm(a,b)\cdot c}{gcd(lcm(a,b), c)}$$ Put the value you obtained already and the it becomes: $$lcm(a,b,c)$$
Hence proving the proposition.
More generally, $$\frac{lcm(a,b)\cdot c}{gcd(lcm(a,b),c)}=lcm(a,b,c)\ \ \ \ \ \forall a,b,c \in Z^+$$