Is $\phi(ab)\ge\phi(a)\cdot \phi(b)$ true for every positive integer pair $(a/b)$? 
If $\phi(n)$ is the totient-function, does $$\phi(ab)\ge \phi(a)\cdot \phi(b)$$ hold for every pair $(a,b)$ of positive integers ? And does equality hold if and only if $\gcd(a,b)=1$ ?

I defined $$g:=\gcd(a,b)$$ $$a':=\frac{a}{g}$$ $$b':=\frac{b}{g}$$ and tried to reduce the problem to the $a'$ and $b'$, but this approach led to nowhere.
 A: Yes.
Using the formula $\phi(ab)=\phi(a)\phi(b)\frac{\gcd(a,b)}{\phi(\gcd(a,b))}$, we can see that
$$\phi(ab) \geq \phi(a)\phi(b) \iff \frac{\gcd(a,b)}{\phi(\gcd(a,b))}\geq1$$
Denoting $c=\gcd(a,b)$, we just need to prove $c\geq \phi(c)$. However, this is always true, since $\phi(n)$ counts the number of positive integers up to $n$ relatively prime time to $n$, which can't ever be greater than $n$. We can also see the only solution to $c=\phi(c)$ is $c=1$, so we have $\phi(ab)=\phi(a)\phi(b) \iff \gcd(a,b) =1$.
A: Let $\{p_i\}$ be the list of primes dividing $\gcd (a,b)$.  Then we can write $$a=\prod p_i^{a_i}\times \prod q_j^{\alpha_j}\quad \&\quad b=\prod p_i^{b_i}\times \prod r_k^{\beta_k}$$
Where the $q_j,r_k$ are primes disjoint from each other and from the $p_i$.
We can now compute both sides of your desired inequality.  We get $$\varphi(ab)=\prod p_i^{a_i+b_i-1}(p_i-1)\times \varphi\left(\prod q_j^{\alpha_j}\right)\times \varphi\left(\prod r_k^{\beta_k}\right)$$
While $$\varphi(a)\varphi(b)=\prod p_i^{a_i+b_i-2}(p_i-1)^2\times \varphi\left(\prod q_j^{\alpha_j}\right)\times \varphi\left(\prod r_k^{\beta_k}\right)$$
From this we see that we can compute the ratio $$\boxed {\frac {\varphi(ab)}{\varphi(a)\varphi(b)}=\prod \frac {p_i}{p_i-1}}$$
The inequality you desire follows at once (as well as the claim that equality requires the gcd to be $1$).
Examples: 
I.    $a=12, b=16$.  Then the only $p_i$ is $2$ and we remark that $$\varphi(192)=64=2\times \varphi(12)\times \varphi(16)$$
II.  $a=18,b=60$.  Then the $p_i$ are $2,3$ and we have $$\frac {\varphi(18\times 60)}{\varphi(18)\times \varphi (60)}=3=\frac 21\times \frac 32$$
III. $a=10,b=45$.  In this case the ratio comes out $\frac 54$ as desired (I've included this examples just to illustrate that, of course, the ratio need not always be an integer).
