More $ϕ(n)$

This solution is based on the assumption that such a composite $n$ exists.

1- Obviously $n\neq p^k*m$ , where $k\geq 2$ and $p$ is a prime and $GCD(p,m)=1$ . Otherwise $p|\phi(n)$ and we know that $p \nmid n-1$ . Therefore, $n$ is a square free composite.

2- If $n=pq$ , where $p$ and $q$ are distinct primes, then

$(p-1)(q-1)|pq-1 \implies p-1|q-1 , q-1|p-1 \implies p=q$

which suggests that $p$ and $q$ are not distinct.

3- If $n=pqr$ , where $p$ , $q$ and $r$ are distinct primes and $p>q>r$ , then, we need $2 \phi(n) \leq n$ and it would be possibly achieved, when $p,q,r$ are as small as possible.

$\phi(n)=(p-1)(q-1)(r-1)> (p-1).4.2=8(p-1) \implies 2.8(p-1)=16(p-1) > 3.5.(p)-1=15p-1 \implies p<16$

one can check all the possible $p,q,r<16$ to realise they won't satisfy the requirement.

4- Note that we need

$\frac{\prod_{j=1}^{k} p_j}{\prod_{j=1}^{k} (p_j-1)} \geq 2 \implies \sum_{j=1}^{k} \log(p_j)- \sum_{j=1}^{k} \log(p_j-1) \approx \sum_{j=1}^{k} \frac{1}{p_j}\geq \log 2$

for some $k \in \mathbb{N}$ .

$\sum_{j=1}^{k} \frac{1}{p_j}$ is a divergent sum, hence, $\frac{\prod_{j=1}^{k} p_j}{\prod_{j=1}^{k} (p_j-1)} \geq 2$ would be satisfied for some $k \in \mathbb{N}$ .