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?
The formula for $\phi(n)$ yields:
$$\phi(n)= n \prod_{p |n} \frac{p-1}{p} $$
Now list the primes dividing n increasing $p_1,...,p_{\omega}$ and use $p_k \geq k+1$.
Then $$ n \prod_{p |n} \frac{p-1}{p} \geq n \prod_{k=1}^\omega (1-\frac{1}{k+1}) $$
and relabel the last product.
The inequality $2^{\omega(n)}\leq \tau(n)$ follows from the definition of $\tau(n)$ and then taking the logarithms of the inequality $2^\omega \leq n$, yields what you need