I think you've gotten a couple of things switched around, or I might be the dyslexic one, I'm not sure. Your example suggests that $\phi(m_1 m_2) > \phi(m_1) \phi(m_2)$. But an example only proves one case out of infinitely many.
You should already know that if $\gcd(m_1, m_2) = 1$, then $\phi(m_1 m_2) = \phi(m_1) \phi(m_2)$. You should also know this formula: $\phi(p^x) = (p - 1)p^{x - 1}$, where $p$ is a prime number and $x$ is a positive integer.
Let's say $m_1 = pq$ (where $q$ is another prime) and $m_2 = pr$ (where $r$ is yet another prime, distinct from $p$ and $q$). Then $\phi(m_1 m_2) = \phi(p^2 qr) = (p^2 - p)(q - 1)(r - 1),$ $\phi(m_1) = (p - 1)(q - 1), \phi(m_2) = (p - 1)(r - 1),$ $\phi(m_1) \phi(m_2) = (p - 1)(q - 1)(p - 1)(r - 1) = (p^2 - 2p + 1)(q - 1)(r - 1).$
In both products we have $(q - 1)(r - 1)$. If we divide that out, we're left with $(p^2 - p) > (p^2 - 2p + 1).$
This still does not prove every possible case specified by your question, but at least it does prove more than just one specific case.