Elementary proof for $\varphi(n)\ge\frac n{2\log(\log(n))}$ Trying to derive an elementary bound for Euler totient function $\varphi(n)$ to be $\mathcal{O}(\frac n{\log(\log(n))})$, I thought to prove a weak version of the well-known inequality
$${\displaystyle \varphi (n)>{\frac {n}{e^{\gamma }\;\log \log n+{\frac {3}{\log \log n}}}}\quad {\text{for }}n>2}$$
for large integers $n\in\mathbb{N}$.
Simplest modification of the theorem would be considering
$$n>e^{e^3} \implies 2\log(\log(n))>e^{\gamma }\;\log \log n+{\frac {3}{\log \log n}}$$
And hence deriving $\varphi(n)>\frac{n}{2\log(\log(n))}$.
I was then stuck trying to prove this.
I've reached bounds of other orders in non-analytic methods (such as $\varphi(n)>\frac{6n}{\pi^2(1+\log(n))}$ and $\varphi(n)>\frac n{4\log(n)}$ ) but I've had no progress trying to prove bounds of this order. (Proving the bound with any other constant would also count, so it's basically enough to prove that $\varphi(n)\ge\frac n{c\log(\log(n))}$ for some $c\in\mathbb{R}^+$ for all $n>N$ where $N$ is a fixed positive integer, $e^{e^3}$ in my case)
Any hints or ideas would be appreciated!
 A: By elementary, I mean "without the use of complex analysis." Using elementary methods (not as deep as PNT), it is possible to show that
$$
\prod_{p\le x}\left(1-\frac1p\right)={e^{-\gamma}\over\log x}\left\{1+O\left(1\over\log x\right)\right\}.
$$
This indicates that when $1\le t\le n$, there is
\begin{aligned}
{\varphi(n)\over n}
&\ge\prod_{p\le t}\left(1-\frac1p\right)\prod_{\substack{p|n\\p>t}}\left(1-\frac1p\right) \\
&\ge{e^{-\gamma}\over\log t}\left(1-\frac1t\right)^s\left\{1+O\left(1\over\log t\right)\right\},
\end{aligned}
where $s$ denotes the number of prime divisors of $n$ that are greater than $t$, which satisfies
$$
s=\sum_{\substack{p|n\\p>t}}1\le\sum_{p|n}{\log p\over\log t}\le{\log n\over\log t}
$$
Thus, we see that when $t=\log n$, there is
$$
-\log\left(1-\frac1t\right)^s=\frac st\left\{1+O\left(\frac1t\right)\right\}\le{\log n\over t\log t}\left\{1+O\left(\frac1t\right)\right\}=O\left(1\over\log\log n\right).
$$
Plugging this result back gives us the lower bound as follows
$$
{\varphi(n)\over n}\ge{e^{-\gamma}\over\log\log n}+O\left(1\over(\log\log n)^2\right).\tag1
$$
To see why this is the best bound possible, set $n_k$ to be the product of the first $k$ primes. Then we have
$$
{\varphi(n_k)\over n_k}=\prod_{1\le i\le k}\left(1-{1\over p_i}\right)={e^{-\gamma}\over\log p_k}\left\{1+O\left(1\over\log p_k\right)\right\}.
$$
Because Chebyshev's bounds ensures that there exists $c_2>c_1>0$ such that $c_1p_k<\log n_k<c_2p_k$ for large $k$, we have
$$
\log\log n_k=\log p_k\left\{1+O\left(1\over\log p_k\right)\right\}=\log p_k\left\{1+O\left(1\over\log\log n_k\right)\right\},
$$
which indicates that (1) holds with equality if $n=n_k$.
