Prove $\frac{pq}{(p-1)(q-1)} < 2$ for distinct odd primes $p,q$ I need this lemma for another proof I'm doing, but I can't crack it. I want something of the structure:
$$\frac{pq}{(p-1)(q-1)} < \dots = \frac{pq}{\frac{1}{2}pq} = 2,$$ 
but I can't figure out what to do with the denominator. 

If $p\le q$ are odd primes, then $p=3+a$ and $q=5+b$ for some $a,b\ge0$.  We have
$$\begin{align}
{pq\over(p-1)(q-1)}={(3+a)(5+b)\over(2+a)(4+b)}\lt2
&\iff15+5a+3b+ab\lt2(8+4a+2b+ab)\\
&\iff0\lt1+3a+b+ab
\end{align}$$
The inequality has very little to do with $p$ and $q$ being either distinct, odd, or prime.  
