Euler's Totient Function: $\phi(n)\geq n\cdot 2^{-r}$.

Question

My friend's teacher made a list with this problem: If $n$ has $r$ distinct prime factors, show that:

$$\phi(n)\geq n\cdot 2^{-r}$$

I tried to help her, but I am not very good in number theory

Answer 1

Hint: If $P$ is prime, then $1-\frac1p\geqslant\frac12$. Therefore, if $p_1,\ldots,p_r$ are prime numbers, then$$\left(1-\frac1{p_1}\right)\left(1-\frac1{p_2}\right)\cdots\left(1-\frac1{p_r}\right)\geqslant 2^{-r}.$$

Answer 2

Hint $$ \phi(n)=n\prod_{i=1}^n\left(1-\frac{1}{p_i}\right) $$ where $n=p_1p_2\dotsb p_r$. Use the fact that $p_i\ge 2$ to bound the product.