TITLE: $\tau(n)\phi(n)\ge n$
QUESTION [6 upvotes]: how to prove $\forall n \in \Bbb N$ $$\tau(n)\phi(n)\ge n$$
$\tau(n)$ is number of positive divisor of $n$
my efford: if $n=p$ is prime then $\tau(p)=2,\phi(p)=p-1,2p-2\ge p$ but how prove for composite number?

REPLY [2 votes]: Do it for powers of primes and the multiplicative property will give you the rest.