We have Andirca's conjecture as: $$\sqrt{p_{n+1}}-\sqrt{p_{n}}<1$$ I have no clue if this is a proof. I was just messing around. (And plus, my methods are probably too elementary). Denote $g(n)$ as the difference between $p_{n+1}$ and $p_n$. i.e., $p_{n+1}=p_n+g(n)$. Playing and messing around a little bit gives us : $$p_{n+1}-2\sqrt{p_{n}p_{n+1}}<1$$ So we have to prove this. In other words, prove that:$$p_{n+1}<2\sqrt{p_{n+1} p_n}+1$$ So we make a claim: Claim: $$p_{n+1}<\sqrt{2 p_{n+1} p_n }$$
Proof: This statement is equivalent to $$(p_{n+1})^2<2(p_{n+1})(p_n)$$ This is quite easy to prove. I am not good with latex. What I did was write $p_{n+1}=p_n+g(n)$. Then expanding both sides and cancelling out gives us: $$g(n)^2<(p_n)^2,$$ which is obvious. Now using this claim and working backwards, we obtain the initial result.
I think I am wrong, but I just don't know where. It seems too elementary.
UPDATE: I have a new claim. Messing around some more, I obtained $$g(n)^2/(4p_n)<1$$ is equivalent to Andrica's conjecture.
UPDATE: I reduced this conjecture to $2S\sqrt{a}>g(n)$, where $S$ is the sum from $n\leq x$ of the $\frac{\mu(n)}{n}$, and $a=p_{n+1}-g(n)$. And since $S=1$, we have $2\sqrt a>g(n)$