Is there a way to show that $\sqrt{p_{n}} < n$?
In this article, I show that $f_{2}(x)=\frac{x}{ln(x)} - \sqrt{x}$ is ascending, for $\forall x\geq e^{2}$. As a result, $\forall n \geq 3$ $$\frac{p_{n}}{ln(p_{n})} - \sqrt{p_{n}}\leq \frac{p_{n+1}}{ln(p_{n+1})} - \sqrt{p_{n+1}}$$ Also (and as a result), $\forall n \geq 3$ $$ \frac{p_{n}}{ln(p_{n})} - \sqrt{p_{n}} > 0$$ Or $$ \frac{\pi (p_{n})}{p_{n}/ln(p_{n})} < \frac{\pi (p_{n})}{\sqrt{p_{n}}}$$
According to PNT $$\displaystyle\smash{\lim_{n \to \infty }}\frac{\pi (p_{n})}{p_{n}/ln(p_{n})}=1$$ Or, $\forall \varepsilon >0$, $\exists N(\varepsilon )$: $\forall n>N(\varepsilon )$ $$1- \varepsilon < \frac{\pi (p_{n})}{p_{n}/ln(p_{n})} < 1+ \varepsilon$$ Or $$1- \varepsilon < \frac{\pi (p_{n})}{p_{n}/ln(p_{n})} < \frac{\pi (p_{n})}{\sqrt{p_{n}}}$$ As a result $\forall \varepsilon >0$, $\exists N(\varepsilon )$: $\forall n>N(\varepsilon )$ $$(1 - \varepsilon ) \cdot \sqrt{p_{n}} < \pi (p_{n}) = n$$
But this is not enough.
Interestingly, Andrica's conjecture is true iff function $f_{4}(x)=\pi (x) - \sqrt{x}$ is strictly ascending ($x < y \Rightarrow f(x) < f(y)$) for prime arguments.
If $f_{4}(p_{n}) < f_{4}(p_{n+1})$ then $$\pi (p_{n}) - \sqrt{p_{n}} < \pi (p_{n+1}) - \sqrt{p_{n+1}}$$ Or $$\sqrt{p_{n+1}} - \sqrt{p_{n}} < \pi (p_{n+1}) - \pi (p_{n}) =1$$
And vice-versa, if $$\sqrt{p_{n+1}} - \sqrt{p_{n}} < 1$$ Then $$-\sqrt{p_{n}} < -\sqrt{p_{n+1}} + 1$$ Or $$\pi (p_{n})-\sqrt{p_{n}} < \pi (p_{n}) + 1 -\sqrt{p_{n+1}} = \pi (p_{n+1}) -\sqrt{p_{n+1}}$$
So, if Andrica's conjecture is true then $\forall n \geq 3$ $$\pi (p_{n})-\sqrt{p_{n}} > 0$$ Or $$\sqrt{p_{n}} < \pi (p_{n})= n$$