Using AM-GM to reason about the upper bound of an $n$th root of a factorial I am working on an alternative argument for the Sylvester-Schur theorem.
I want to show that for $n > 18$, $n+1 > \sqrt[\leftroot{-1}\uproot{2}\scriptstyle \left(n - \pi(n) - 1\right)]{ n! }$.
Can be AM-GM be used to demonstrate this?
Here's my thinking:
(1) For $n \ge 2, \pi(n) \le \frac{n}{3} + 2$

The prime counting function $\pi(n) \le \frac{n}{3} + 2$ since for a prime $p > 3, p \equiv 1 \pmod 6$ or $p \equiv 5 \pmod 6$ so that $\pi(n) \le 2 + \left\lfloor\frac{n+1}{6}\right\rfloor + \left\lfloor\frac{n+5}{6}\right\rfloor - 1 \le \frac{2n+12}{6} = \frac{n+6}{3}$.

(2) For $n > 18, \frac{n-2}{2} > \pi(n)$ 

If $n > 18$, then $\frac{n-2}{2} > \frac{n}{3}+2 \ge \pi(n)$ since $3n - 6 > 2n+12$.

(3) $1 > \dfrac{n}{2(n-\pi(n) -1)}$

This follows from $2(n - \pi(n) -1) > n$ since $n-2 > 2\pi(n)$.

(4) Multiplying $n+1$ to both sides of the equation:
$$n+1 > \frac{n(n+1)}{2(n-\pi(n)-1)}$$
Can I now use AM-GM and the summation formula of $\sum\limits_{i \le n}{i} = \frac{(n+1)(n)}{2}$ to conclude:
$$n+1 > \frac{n(n+1)}{2(n - \pi(n)-1)} > \sqrt[\leftroot{-1}\uproot{2}\scriptstyle \left(n-\pi(n)-1\right)]{n!}$$       
 A: Strictly speaking, $\sqrt[k]{n!} \not \leq \frac{n(n+1)}{2k}$ for all $k \geq 1$. (The Left Hand Side converges to 1 and the right converges to 0 as $k \to \infty.)
The inequality fails for $n = 10$, where $\pi(10) = 4$ and so $$\frac{10(11)}{2(10-4-1)} = 11 < 20.5093835306 \approx \sqrt[10-4-1]{10!}$$ and for $n = 19$, $\pi(19) = 8$, so $$\frac{19(20)}{2(19-8-1)} = 19  < 51.1104214517 \approx \sqrt[10]{19!}$$
Even for $n = 100$, $\pi(100) = 25$ (pi(n)) so
$$\frac{100(101)}{2(100-25-1)} = 68.2432432432 < 136.373434808 \approx \sqrt[10-25-1]{100!}$$
So your inequality fails for $n$ as large as $100$.(You can check my work, if you want.)
The only redeeming property seems to be that for $k_n = n-\pi(n) -1 < n$, $\frac{Cn}{\log(n)}\leq \pi(n) \leq \frac{Dn}{\log(n)}$ so $k_n$ is slightly smaller than $1/n$. See (1) below. Even without that, $k_n > 0$, as you showed above.
A direct approach does not seem to work (but there might be others), because:
$$(n!)^{1/k_n} = (n!)^{1/n}(n!)^{1/k_n - 1/n} < \frac{n(n+1)}{2n}(n!)^{(\pi(n)+1)/(n(n-\pi(n)-1))} = \frac{n(n+1)}{2(n-\pi(n)-1)} \left(\frac{n-\pi(n)-1}{n}\right)(n!)^{(\pi(n)+1)/(n(n-\pi(n)-1))} < \frac{n(n+1)}{2(n-\pi(n)-1)} (n!)^{(\pi(n)+1)/(n(n-\pi(n)-1))}.$$
The question is how to control that exponent. which you need to be less than $1/n$, because you need the factorial term to vanish. If you use the estimates for $\pi(n)$, then you can get that it is at most
$$\frac{Dn/\log(n)+1}{n(n-Cn/\log(n)-1)} =  \frac{D/\log(n)+1/n}{n(1-C/\log(n)-1/n)}$$
which is at most $Const.\left(\frac{1}{n\log(n)}\right)$. But since $\sqrt[n]{n!} \sim n/e$ by Stirling's Inequality and $n^{1/\log(n)} = e$, (See here) we see that this term is just bounded by a constant. So, at the end of the day what I got for large $n$ by applying AM-GM in "the obvious way" was $(n!)^{1/k_n} < Const. \frac{n(n+1)}{2(n-\pi(n)-1)}$
(1) $C, D > 0$ exist (perhaps for large $n$) according to an "elementary" proof (meaning just combinatorics, not complex analysis or such)  in Number Theory - Andrews, (I don't have the book with me) and $C = 1, D = 1.25506$ according to Prime Counting
