How can I prove this inequality? $\frac{p_k}{p_{k+1} + 1} < \frac{\ln(\ln(Primorial(k)))}{\ln(\ln(Primorial(k+1)))}$ How can I prove the following inequality? Or what are some tips that would help me prove it myself?
$$\frac{p_k}{p_{k+1} + 1} < \frac{\ln(\ln(Primorial(k)))}{\ln(\ln(Primorial(k+1)))}$$     for $k >= 3$, where $p_k$ is the $k$th prime, and $Primorial(k)$ is the product of the first $k$ primes
If we can't prove it for all $k >= 3$, can we at least prove it is assymptotically true?
 A: Not an answer, more like two long comments ..

Function $f(x)=\frac{x}{\ln{x}}$ is ascending for large $x$'s (easy to show, check the $1$st derivative). This means that for large enough $n$'s
$$\frac{p_n}{\ln{p_n}}\leq \frac{p_{n+1}+1}{\ln{(p_{n+1}+1)}}$$
Now:
$$\frac{p_{n}}{p_{n+1}+1}=
\frac{p_{n}}{\ln{p_{n}}}
\cdot\frac{\ln({p_{n+1}+1)}}{p_{n+1}+1}
\cdot\frac{\ln{p_{n}}}{\ln({p_{n+1}+1)}}\leq \frac{\ln{p_{n}}}{\ln{(p_{n+1}}+1)} \tag{1}$$

Chebyshev function
$$\vartheta (p_{n})=\sum\limits_{k=1}^n\ln{p_k}=\ln{(\operatorname{Primorial}(n))}$$
has the property (section 1.1) that
$$\lim\limits_{n\to\infty}\frac{\vartheta (p_{n})}{p_n}=1$$
which means that for large enough $n$'s
$$p_n(1-\varepsilon)<\vartheta (p_{n}) < p_n(1+\varepsilon) \Rightarrow\\
\ln{p_n}+\ln{(1-\varepsilon)}<\ln{(\ln{(\operatorname{Primorial}(n))})} < \ln{p_n}+\ln{(1+\varepsilon)}$$
and
$$\frac{\ln{p_n}+\ln{(1-\varepsilon)}}{\ln{p_{n+1}}+\ln{(1+\varepsilon)}}<
\frac{\ln{(\ln{(\operatorname{Primorial}(n))})}}{\ln{(\ln{(\operatorname{Primorial}(n+1))})}}<
\frac{\ln{p_n}+\ln{(1+\varepsilon)}}{\ln{p_{n+1}}+\ln{(1-\varepsilon)}}$$
or
$$\frac{\ln{(\ln{(\operatorname{Primorial}(n))})}}{\ln{(\ln{(\operatorname{Primorial}(n+1))})}} \sim \frac{\ln{p_n}}{\ln{p_{n+1}}} \tag{2}$$

Now compare $(1)$ and $(2)$ ...
