Let $p_n$ denote the $n$th prime. Recall Andrica's conjecture, which states that $$\sqrt{p_{n+1}}-\sqrt{p_n}<1\quad\text{ for all }\,n.$$ I think Andrica's conjecture implies Bertrand's postulate. I have found that it actually implies $\frac{p_{n+1}}{p_n} < \frac32$ (for $p_n>11$). Is this true?
Assume $$(\sqrt{p_{n+1}}-\sqrt{p_n})(\sqrt{p_{n+1}}+\sqrt{p_{n}}) > \frac{p_{n}}2$$ ( in other words $p_{n+1}/p_{n} > 3/2$). Assume also $$1> \sqrt{p_{n+1}}-\sqrt{p_{n}}$$ (this is Andrica's conjecture) then $\sqrt{p_{n+1}}+ \sqrt{p_{n}} > p_{n}/2$ therefore $$\frac{\sqrt{p_{n+1}}}{\sqrt{p_{n}}} + 1 > \frac{\sqrt{p_{n}}}2.$$ So $\sqrt2+ 1 > \sqrt{p_{n}}/2$, this implies $24 > p_{n}$. Note, $p_{n}$ means the $n$-th prime.
So if Andrica's conjecture is true for any $p_{n} > 24$ then $$\frac{p_{n+1}}{p_{n}} <\frac32,$$ for any primes greater than 29.
If $p_{n+1}-p_n> p_n/2$. $$(\sqrt{p_{n+1}}-\sqrt{p_n})(\sqrt{p_{n+1}}+\sqrt{p_n})>\frac{p_n}2$$ If $\sqrt{p_{n+1}}-\sqrt{p_n} <1$ therefore $\sqrt{p_{n+1}}+\sqrt{p_n}>p_n/2$.
Therefore $\sqrt{p_{n+1}/p_n}+1 > \sqrt{p_n}/2$, therefore $\sqrt2 + 1 > \sqrt{p_n}/2$ hence $24>p_n$.
So if $\frac{p_{n+1}}{p_n} > \frac32$ and $\sqrt{p_{n+1}} - \sqrt{p_n} <1$ then $p_n < 24$.
Therefore if $\sqrt{p_{n+1}} - \sqrt{p_n} <1$ and $P_n > 24$ then $\frac{p_{n+1}}{p_n} < \frac32$. (Note if $p_n >2 p_{n+1}/p_n$ does not equal $3/2$.)
Several conjectures (and results) on prime numbers can be expressed using gap size $$g_n=p_{n+1}-p_n.$$
Bertrand's postulate is equivalent to $g_n<p_n$.
Oppermann's conjecture is equivalent to $g_n<\sqrt{p_n}$
Andrica's conjecture says that $\sqrt{p_{n+1}}-\sqrt{p_n}<1$. If we multiply both sides of the inequality by $\sqrt{p_{n+1}}+\sqrt{p_n}$ we get $$g_n=p_{n+1}-p_n<\sqrt{p_{n+1}}+\sqrt{p_n}.$$ Using $\sqrt{p_{n+1}}-\sqrt{p_n}<1$ once again we have $\sqrt{p_{n+1}}<\sqrt{p_n}+1$ and $$g_n < 2\sqrt{p_n}+1.$$
It is clear that Opperman's conjecture implies both Bertrand's postulate and Andrica's conjecture
Since $2\sqrt{p_n}+1<p_n$ for $p_n\ge11$, Andrica's conjecture implies Bertrand's postulate.
And also for any positive constant $c$ we have $$2\sqrt{p_n}+1< c p_n$$ for all large enough $n$'s.
So this implies $g_n < cp_n$, i.e. $$p_{n+1}<(1+c)p_n$$ for large enough $n$. (Which means that there is only finitely many $n$'s left to check.)