Solution by numbersandnumbers. Clearly $p(n)$ is just the partition number of $n$. We will prove the equivalent inequality $p(0) +p(1)+... + p(n-1) \le p(n)\sqrt{2n}$. Notice that on the LHS, $p(n-i)$ for each $i$ is equivalent to the number of partitions of $n$ with at least one element equal to $i$. Therefore, by a simple double count, $p(0) + p(1) + ... + p(n-1) = \displaystyle\sum_{p \in S} f(p)$, where $S$ is the set of all partitions of $n$ and $f(p)$ denotes the number of distinct elements of $p$. However, it's clear that for each $p$ we have $f(p) \le \sqrt{2n}$, since $1 + 2 + ... + \lceil \sqrt{2n} \rceil >n$, so $\displaystyle\sum_{p\in S}f(p) \le p(n)\sqrt{2n}$ as desired.

A link to a solution by Noam Elkies, at mathstackexchange: https://math.stackexchange.com/questions/1357982/on-the-inequality-frac1p1p2-dots-pn-1pn-leq-sqrt-2n Original Source. The original source seems to be 1984 Tournament of Towns.