Proof of Andrica when Assuming Oppermann Proof of Andrica's conjecture by assuming Oppermann's conjecture. 
Oppermann's conjecture:
$$n\geq2\wedge\pi\left(n^{2}-n\right) < \pi\left(n^{2}\right) < \pi\left(n^{2}+n\right).$$  
Andrica's conjecture:
$$i\geq2\wedge\sqrt{p_{i+1}}-\sqrt{p_{i}}<1.$$
Insert Gottfried Helms' nice proof here$\ \ \Box$
 A: [update]: Upps, I see that while I'm just writing @Dan has undeleted his earlier answer. So priority (and thus also the bounty) is at Dan's, of course

I) Let's look at some examples due to Oppermann's consideration:
$$ \small \begin{array}{}  m_{13}=13^2=169 & a_{13} = 169-13+1=157 & b_{13}=169+13-1 = 181 \\
  m_{14}=14^2=196 & a_{14} = 196-14+1=183 & b_{14}=196+14-1 = 209 \\
  m_{15}=15^2=225 & a_{15} = 225-15+1=211 & b_{15}=225+15-1 = 239 \\ 
\end{array}$$
or , written more compactly
$$ \small \begin{array}{} 
  m_{13}: &   157...168 & 169 &170... 181 \\
  m_{14}: & 183...195 & 196 & 197... 209 \\
  m_{15}: & 211...224 & 225 & 226...239 \\ 
\end{array}$$
We can see that the intervals for consecutive $m_k$ cover the integers except the numbers $n^2$, $n^2+n$ etc, which by definition are not prime.
Oppermann's conjecture is true, if in each of the intervals is at least one prime.        

II) But because the intervals follow each other immediately, we can also look at the two intervals between the two consecutive squares $n^2$ and $(n+1)^2$. 
$$ \small \begin{array} {rrrrr}
 n^2 &  \big[(n^2+1) \cdots (n^2+n-1)\big]  & (n^2+n) & \big[(n^2+n+1) \cdots ((n+1)^2-1) \big] & (n+1)^2 \\
\end{array}$$
The endpoints $n^2$ , $(n+1)^2$ and the center $n^2+n = (n+1)^2-(n+1)$ are not prime by construction, so the maximal distance between two primes can - assuming Oppermann - at most be $((n+1)^2-1) - (n^2+1)$.       
III) Now express the inner neighbours of the two endpoints as $p_0 = n^2+1$ and $p_1=(n+1)^2-1$ which mark the greatest possible distance between two numbers in that interval , then we can refer to the Andrica-formulation $\sqrt{p_1} - \sqrt{p_0}<1$ as
$$ \sqrt{((n+1)^2-1)} - \sqrt{(n^2+1)}\lt 1$$
But this is with a small $\small 0 \le \delta \lt 1$ and thus $\small \sqrt{n^2+1} \sim n+\delta $ and $ \small \sqrt{(n+1)^2-1} \sim (n+1)-\delta $
$$ \big((n+1)-\delta \big)  - \big(n +\delta \big) = 1 - 2 \delta \lt 1$$
and thus always (for $n$ greater than some small value, perhaps even if greater than 1) true.         
IV) From this follows, that Oppermann's conjecture implies Andrica's conjecture
