I reformulated the question by entering my demonstration attempt
$$\prod_{2\lt p \le\sqrt 2n}\left(\frac{p-1}{p-2}\right)^2\le\frac n2$$
I tried to prove it this way: Let n be an integer, p be a prime. We have a chain of equivalences: $$\prod_{i=2}^{\pi(\sqrt 2n)}\left(\frac{pi-1}{pi-2}\right)^2\le\frac n2$$ $$\prod_{i=2}^k\left(\frac{pi-1}{pi-2}\right)^2\le\frac n2$$ , for $$ \pi(\sqrt 2n)=k $$ But it is an easy task to prove $$ 2 \phi(\theta_{k}) \lt P_{k} \prod_{i=2}^k(Pi-2)$$ $$\prod_{2\lt p \le\sqrt 2n}\left(\frac{p-1}{p-2}\right)^2\le\frac n2$$ where $\theta k$=3 . 5 .7 ....P_{k}
Indreed $$4 \phi^2(\theta_{k})\lt Pk^2 \prod_{i=2}^k(Pi-2)^2$$ $$ 2\phi(\theta k)\lt P_{k} \prod_{i=2}^k(Pi-2)$$
Our next task is to prove that $$ 2\phi(\theta_{k})\lt \phi $$ $$\phi\lt P_{k} \psi_{k} $$ wher $$\psi_{k} = \prod_{i=2}^k(Pi-2)$$ By inspection we can verify that for $$K \le 4$$ $$ 2\phi(\theta_{k})\lt \theta_{k} $$ We have left out to prove that $$K \le 4$$ $$\theta \lt P_{k} \psi $$ Assume at the contrary that $$ P_{k} \psi _{k}\le\theta _{k}$$ We have $$ P_{k} \psi_{k}\le\theta_{k}$$ $$\psi_{k} \le \theta_{k-1} $$ $$ \prod_{i=2}^k(Pi-2) \lt 3 .5 .7 ....P_{k}-1$$
Now it is clair that the last relation is patently false..
According to you is correct?
Thank's