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!

• See here. It relies on Mertens' theorems, but are not exactly elementary due to their relations to PNT. Jul 28, 2022 at 8:14
• I've checked it out before but I need something that fully relies on elementary number theory. Merten's theorem itself is not an elementary theorem Jul 28, 2022 at 9:36
• @TheSimpliFire Mertens' theorems are themselves elementary, and proofs of them do not require PNT. Jul 28, 2022 at 15:22

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 $$c_1p_k<\log n_k for large $$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$$.
• @GregMartin I guess you got it the wrong way. It doesn't matter if the solution is complicated, I just need it to be "elementary" (not using analytic/algebraic number theory methods) Let's say I have a function $f(n)=g(n)\cdot\varphi(n)$ and I need to find a positive integer $n_0$ such that $f(n)>k$ for all $n\ge n_0$ where $k$ is an arbitrary integer. And I got a non-asymptotic lower bound for $g(n)$. That's why I need a non-asymptotic lower bound for totient function. Jul 29, 2022 at 21:53