Let $\omega$ be the number of distinct primes dividing n, then $\varphi(n)\geq n\prod_{k=2}^{\omega(n)+1}\left(1-\frac{1}{k}\right)=\frac{n}{\omega(n)+1}$ Also $2^{\omega(n)}\leq \tau(n) \leq n$, and conclude that $\varphi(n)\geq \frac{cn}{\log n} ,n\geq2$ for a suitable constant $c>0.$
$\tau(n)$ denotes the number of positive divisors of $n$.
Could some give me a proof?