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. 

Assume $p>q$:
$$
2(p-1)(q-1)-pq=pq-2p-2q+2>pq-4p+2=p(q-4)+2
$$
which is $>0$ for $q>3$. For $q=3$,
$$
2(p-1)(q-1)-pq=4(p-1)-3p=p-4>0
$$
Therefore $2(p-1)(q-1)>pq$.
