Prime number inequality Hey I recently got interested in number theory and proved the following inequality:
\begin{align}
    x^m-y^m >  p_n^2 + 4p_n + 3 > 1\ \text{(corrected again)}                   
  \end{align}
where 
$x-y\neq1$ and m in an integer >1
and
\begin{align}
\gcd(x,y)&=1  \\
    xy&= 2*3*5*..*p_n                         
  \end{align}
So my question is ... Does this formula already exist? And is it useful or slightly interesting?
 A: The inequality deals with the primorial function
$$
p_n\#=2\cdot3\cdot5\cdots p_n=\prod_{p\le p_n,\ p\in \mathbb{P}}p,
$$
where $p_n$ is the $n$th prime.
Asymptotically we have the result that
$$
\lim_{n\to\infty}\frac{\ln p_n\#}{p_n}=1.
$$
Early on the primorials are bit smaller though. For example $\ln(59\#)\approx49$.
Consider the following problem. Assume that $xy=p_n\#$ and that $x-y\ge3$ (in OP $x,y$ were constrained to be integers of opposite parity such that $x-y>1$ implying that $x-y\ge 3$). Therefore
$$x^m-y^m\ge x^2-y^2=(x-y)(x+y)\ge3(x+y).$$
Here by the AM-GM inequality $x+y\ge2\sqrt{xy}=2\sqrt{p_n\#}.$ Therefore asymptotically we get a lower bound
$$
x^m-y^m\ge 6\sqrt{p_n\#}\ge6e^{\frac n2(1+o(1))}.
$$
Asymptotically we also have have $p_n\approx n\ln n.$ This suggests that
$$
\frac{\ln(x^m-y^m)}{\ln p_n}\ge \frac n{2\ln n} K(n),
$$
where $K(n)$ is some correction factor (bounded away from zero) that I won't calculate.
Your result says (using only the main term $p_n^2$) that
$$
\frac{\ln(x^m-y^m)}{\ln p_n}\ge 2.
$$
So asymptotically it is weaker. But it would not be fair to call your result trivial because of this. I'm not a number theorist, but I have seen simpler estimates being derived in many number theory books, and in addition to being fun, they pave the road to stronger results.
Please share details of your argument with us, so that we can comment and give you other kind of feedback!
A: Modulo small values of $n,$ you do not need to assume that $x-y\ne 1$ as well as asymptotic on the primorial function. Instead, you can get away with Bertrand's postulate stating that $p_{n}<2p_{n-1}$ or $p_{n-1}\ge p_n/2.$ Indeed, if $m\ge 2,$ then 
$$x^m-y^m=(x-y)(x^{m-1}+...+y^{m-1})\ge (x-y)(x+y)\ge x+y\ge 2\sqrt{xy}.$$
Note, $$2\sqrt{xy}=2\sqrt{p_1...p_{n-3}p_{n-2}p_{n-1}p_n}$$ and using the fact that $p_{n-2}\ge p_n/4,$ $p_{n-3}\ge p_n/8$ and $p_{n-1}\ge p_n/2$ we can estimate
$2\sqrt{xy}\ge 2\sqrt{p_1p_2\cdot...\frac{p_n^4}{64}}=\frac{p_n^2}{8}\sqrt{p_1p_2...}.$ So we are left to check the result for small values of $p_n.$
