Prove $\frac{pq}{(p-1)(q-1)} < 2$ for distinct odd primes $p,q$

Question

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.

Answer 1

As $p\geq3$, $q\geq5$, we have $p/(p-1)\leq3/2$, $q/(q-1)\leq5/4$.

Answer 2

You could prove that $1<\frac{k}{k-1} < \frac{\ell}{\ell-1}$ for $k>\ell$ and then observe that $\frac {3\times 5}{2\times4}<2$, and that choosing larger odd primes will thus make the result smaller.

Answer 3

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 $2(p-1)(q-1)>pq$.

Answer 4

$$\frac{pq}{(p-1)(q-1)} <2\Longleftrightarrow pq-2p-2q+2 =(p-2)(q-2)-2>0$$ for $p\ge 3$ and $q\ge 4$.

Answer 5

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.