# $\forall n,m \in \mathbb{Z}^+ \ \exists p \in \mathbb{Z}^+ \ \ \text{such that} \ \ \frac{1}{m} + \frac{1}{n} > \frac{1}{p}$

Rearranging the required to prove, it's equivalent to
$$\forall n \in \mathbb{Z}^+ \ \ \forall m \in \mathbb{Z}^+ \ \ \exists p \in \mathbb{Z}^+ \ \ \text{such that} \ \ p > \frac{mn}{m+n}.$$

I'm not sure if I'm misinterpreting this question. It sounds like I just need to find some positive integer that gives is of higher value than $\frac{mn}{m+n}$.

This seems like it's always possible due to the integers extending to infinity.
Could I just say that given any $m,n \in \mathbb{Z}^+$, an integer $p$ exists because I can simply choose $p:= \lceil \frac{mn}{m+n}\rceil + k$ for any $k \in \mathbb{Z}^+$?
e.g. Given $m=2,n=3$, $\frac{mn}{m+n} = 1.2$. Since $1.2 < \lceil 1.2 \rceil = 2 \in \mathbb{Z}^+$ then I can pick $p:= 2$.

• Not following. Why not just take $p=n$? – lulu Aug 10 '17 at 1:27
• I think you must have misunderstood or mis-transcribed the question. As it stands, it's true for trivial reasons. – lulu Aug 10 '17 at 1:32
• You are correct. This does seem pretty straightforward. Let $p = \max (m,n)$ then $\frac 1p = \min (\frac 1m, \frac 1n)< 2*\min(\frac 1m, \frac 1n) \le \frac 1m + \frac 1n$. – fleablood Aug 10 '17 at 1:38