Inequality $\varphi(N)>\pi(N)$?

Question

Is it trivial that $\varphi(N)>\pi(N)$ for sufficiently big integers $N$, where $\varphi$ is Euler's totient function and $\pi$ is the prime-counting function?

The only exceptions less than $1.000.000$ are $1,2,3,4,6,8,10,12,14,18,20,24,30,42,60,90$.

Why should the number of to $N$ relative prime numbers less than $N$ be greater than the number of primes less than $N$?

Can someone please hint on this?

Answer 1

Since: $$\phi(N)=N\prod_{p\mid N}\left(1-\frac{1}{p}\right) $$ we have: $$ \frac{N}{\phi(N)}=\exp\left(\sum_{p\mid N}-\log\left(1-\frac{1}{p}\right)\right)\leq K\exp\sum_{p\mid N}\frac{1}{p}$$ while $$\pi(N)\leq \log 4\cdot\frac{N}{\log N}$$ hence we just need to show that for any $N$ big enough: $$\exp\sum_{p\mid N}\frac{1}{p}\leq\frac{\log 4}{K}\log N$$ or: $$ \sum_{p\mid N}\frac{1}{p}\leq C +\log\log N $$ that is trivial since the LHS behaves like $\log\log\log N$.

Answer 2

Hint:-

Use $\varphi(n) >\dfrac{n}{e^\gamma \ln (\ln n)+\dfrac{3}{\ln (\ln n)}}$ and $\dfrac{n}{\ln n -(1+\epsilon)}>\pi(n)$ for all sufficiently large $n$ and for all $\epsilon>0$.