Prove that $\left(\prod\limits_{2Conjecture:
$$\left(\prod_{2<p}^{p_i} \frac{p-1}{p}\right) \cdot \left(p_{i + 1}^2 - p_i^2 \right) > \pi(p_{i + 1}^2) - \pi(p_i ^2) \tag{1}$$
The LHS expression is the product of Mertens's third theorem truncated at $p_i$ (subsequently just $M_x$ for truncation at $x$) times the length of the interval between the squares of $p_i$ and $p_i+1$, and the RHS is the actual number of primes between those squares. For instance, for $p_i = 7$, the expression is:
$$\frac{8}{35} \cdot \left(11^2 - 7^2\right) > \pi(121) - \pi(49)$$
And in fact $16.46 > 15$. Inequality $(1)$ is (perhaps surprisingly) empirically true for all $p_i$. The inequality seems intuitively likely to be true, as we would expect the result of $M_xx^2$ to be larger than $\pi(x^2)$ simply because $e^{-\gamma} > \frac12$ and therefore
$$\pi(x^2) \sim \frac{x^2}{\log x^2} = \frac12\frac{x^2}{\log x} < e^{-\gamma}\frac{x^2}{\log x} \tag{2}$$
(Inequality $(2)$ is empirically true only above $x \approx 100$, because the difference between $M_x \log x$ and $e^{-\gamma}$ is large for small $x$)
We can show (much more easily than I expected, in this question where I briefly failed basic algebra) that the density of primes between $p_{i-1}^2$ and $p_i^2$ is lower than the density of primes between $0$ and $p_i^2$, that is:
$$\frac{\pi\left(p_{i+1}^2) - \pi(p_i^2\right)}{p_{i+1}^2 - p_i^2} < \frac{\pi(p_i^2)}{p_i^2} \tag{3}$$
With this, we expect difference between the LHS and RHS of $(1)$ to be greater than $(2)$ alone would imply.
But... I'm roughly 110% certain that $(2)$ and $(3)$ together don't qualify as proof. They might be proof of asymptotic behavior as $p_i \to \infty$, but I'm wondering if this can be (or already has been!) proven for smaller values. Any links or thoughts from the community?
 A: $\prod \limits_{2\leq p \leq p_i } (1-\frac{1}{p})  (p_{i+1}^2-p_{i}^2) \geq li(p_{i+1}^2)-li(p_{i}^2)$ for all $i \geq 4$
By noticing that $ \prod \limits_{2\leq p \leq p_i } (1-\frac{1}{p})  (p_{i+1}^2-p_{i}^2)  \approx \frac{e^{-\gamma}}{\ln p_i} (p_{i+1}^2-p_{i}^2)\geq  \frac{p_{i+1}^2-p_{i}^2}{2\ln p_i}\geq \int \limits_{p_{i}^2}^{p_{i+1}^2} \frac{dt}{\ln t}=li(p_{i+1}^2)-li(p_{i}^2)$
If you assume R.H. and let $ p_{i+i^{\epsilon}}^2-p_{i}^2$ with $i^{\epsilon} = O(\ln^k i)$ for $ k \geq 2$ then since $ |\pi(x)- li(x)| \leq \sqrt{x} \ln x$ this would imply the correctness of your conjecture for all $ i\geq 3000$ give or take according to the choice of $k$.
But without R.H. i don't see a way.
A: $(2)$ and $(3)$ can be combined to form a proof of the "for large enough $i$'s" style and everything else can be checked with a computer program.
$$\frac{\pi\left(p_{i+1}^2) - \pi(p_i^2\right)}{p_{i+1}^2 - p_i^2} < 
\frac{\pi(p_i^2)}{p_i^2}=
\frac{\pi(p_i^2)\cdot \log{p_i^2}}{p_i^2}\cdot \frac{1}{\log{p_i^2}}\leq ...$$
from PNT, for large enough $i$'s we have
$\frac{\pi(p_i^2)\cdot \log{p_i^2}}{p_i^2}\leq 1+\varepsilon$, thus
$$...\leq \frac{1+\varepsilon}{2\log{p_i}}=
\frac{e^{\gamma}(1+\varepsilon)}{2(1-\varepsilon)}\cdot \frac{e^{-\gamma}\cdot (1-\varepsilon)}{\log{p_i}}\leq ...$$
from Mertens' 3rd theorem, for large enough $i$'s we have
$\prod\limits_{p_i\leq p}\frac{p-1}{p}\geq\frac{e^{-\gamma}\cdot(1-\varepsilon)}{\log{p_i}}$, thus
$$...\leq \frac{e^{\gamma}(1+\varepsilon)}{2(1-\varepsilon)}\cdot\left(\prod\limits_{p_i\leq p}\frac{p-1}{p}\right) \tag{4}$$
What is left is to find a suitable $\varepsilon>0$ such that
$$\frac{e^{\gamma}(1+\varepsilon)}{2(1-\varepsilon)}<1 \iff
1+\varepsilon < \frac{2}{e^{\gamma}}\cdot(1-\varepsilon)$$
which works for $e\leq0.01$ (it's easy to see that $f(x)=\frac{1+x}{1-x}$ is ascending).
