Is the first term of this sequence the greatest? Let $(u_{n})_{n\geqslant 0}$ the sequence defined by $\displaystyle{u_{n}=\dfrac{2}{\log^{2}\dfrac{p_{n+2}^{2}+3}{2}}\prod_{i=2}^{n+2}\dfrac{p_{i}}{p_{i}-2}}$ where $p_{k}$ is the $k$-th prime number.
Is it true that $u_{0}$ is the greatest term of this sequence?
Thanks in advance.
 A: The claim is plausible. I will show that
$$\lim_{n \to \infty} \frac{2}{\log^2 \frac{p_{n+2}^2+2}{3}} \prod_{i=2}^{n+2} \frac{p_i}{p_i-2}\approx  0.600 < u_1 \approx 1.436$$
Set $P = p_{n+2}$. We focus on computing the product
$$X:= \prod_{3 \leq p \leq P} \frac{p}{p-2}$$
We have
$$X = \prod_{3 \leq p \leq P} \left( \frac{(1-1/p)^2}{1-2/p} \right) \cdot \prod_{3 \leq p \leq P} \left( 1-\frac{1}{p}\vphantom{\frac{(1-1/p)^2}{1-2/p}} \right)^{-2}$$
The product $\prod_{p \geq 3} \left( \frac{(1-1/p)^2}{1-2/p} \right)$ is convergent; I don't know a closed form for it but numerical computation suggests that is roughly $1.514$. By Mertens' theorem, the second term is $(1/4) (e^{\gamma} \log P)^2 \approx 0.793 \log^2 P$. (The $1/4$ is because of the missing term at $p=2$.) So
$$X \approx 1.514 \times 0.793 \log^2 P \approx 1.201 \log^2 P.$$
Now 
$$\frac{2}{\log^2\frac{P^2+2}{3}} = \frac{2}{\left( 2 \log P + O(1) \right)^2} \sim \frac{1}{2 \log^2 P}.$$
So your ratio is
$$\frac{2}{\log^2\frac{P^2+2}{3}} X \approx \frac{1}{2 \log^2 P} \left(1.201 \log^2 P \right) \approx 0.600$$
There should be no obstacle in principle to turning these estimates into precise bounds which would allow you to prove that this is true for all $n$ past a finite bound, and then check this finite number of cases by hand. 
