$\tau(n)\phi(n)\ge n$

Question

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?

Answer 1

Do it for powers of primes and the multiplicative property will give you the rest.