At the heart of the standard proof for Betrand's Postulate is the inequality that for $n \ge 41$:
$$\dfrac{2^{\lfloor\sqrt{2n}-3\rfloor}}{n} \ge 1$$
It occurred to me that this can be proven without use of derivatives. I wanted to check if my reasoning is valid or if there is an even easier algebraic argument.
(1) For $w \ge 6$, $2^w > \dfrac{w^2 +8w +16}{2}$ since:
Base Case: $2^7 = 128 > 36+48+16 = 100$
Assume $2^{w+1} > w^2+8w+16$
$2^{w+2} > 2w^2 + 16w + 32 > (w+1)^2 + 8(w+1)+16 = w^2 + 10w +26$
(2) It follows that $n \ge 41$, $2^{\lfloor\sqrt{2n}-3\rfloor} \ge n$ since:
Let $w = \lfloor\sqrt{2n} - 3\rfloor$ so that $w \ge 6$
$\sqrt{2n} - 3< w+1$ so that $n < \dfrac{w^2 + 8w + 16}{2} < 2^w.$