If $n>1$ has $r$ different prime factors, then the totient is bounded by $\varphi(n) \geqslant n/2^r$?

Question

I want to prove that if $n>1$ has $r$ different prime factors, then

$$\varphi(n) \geqslant \frac{n}{2^r}.$$

Answer 1

We have $\phi(n)=n\cdot\prod_{p\mid n}\left(1-\frac1p\right)$ and $1-\frac1p\ge \frac12$.

Answer 2

Use the following: if the prime decomposition of $\;n\in\Bbb N\;$ is

$$n=\prod_{k=1}^m p_k^{a_k}\;,\;\;a_k\in\Bbb N\implies\varphi(n)=n\prod_{k=1}^m\left(1-\frac1{p_k}\right)$$

and thus:

$$m=r\implies \varphi(n)=n\prod_{k=1}^{m=r}\left(1-\frac1{p_k}\right)\ge n\prod_{k=1}^{m=r}\left(1-\frac1{2}\right)=\frac n{2^r}$$