I am trying to understand the proof of a theorem and I am stuck in a specific part. Namely, the following inequality:
$$\frac{d^t}{n^{t-1}} - \binom{n}{r} \bigg(\frac{a+b}{n}\bigg)^r \geq (2c)^r - \frac{n^r}{r!}\bigg(\frac{a+b}{n}\bigg)^r \geq (2c)^r - \bigg(\frac{e(a+b)}{r}\bigg)^r \geq c^r \geq a$$
The information the theorem gives us is that $d\geq 2cn^{1-\frac{1}{r}}$, that $c=\max\{a^{\frac{1}{r}}, \frac{3(a+b)}{r}\}$ and that $r! \geq (\frac{r}{e})^r$, but I cannot for the life of me figure out how each thing is used, so that I can completely understand every step of the inequalities.
If anyone could help I would greatly appreciate it.
Thanks in advance.
I shall use some default assumptions (for instance that $n\ge r$ are positive integers or $a, b\ge 0$). I guess $t$ is $r$, because $d\geq 2cn^{1-\frac{1}{r}}$ perfectly fits to $\frac{d^t}{n^{t-1}}\ge (2c)^r$ and keys “t” and “r” are neighbors. :- ) Next, $$-\binom{n}{r}=-\frac{n(n-1)\dots(n-r+1)}{r!}\ge -\frac{n^r}{r!}.$$ Next, $-\frac 1{r!}\ge –\left(\frac er\right)^r$, because $r!\ge \left(\frac re\right)^r $ (I recall that according to Stirling’s formula, there exists a number $0<\theta<1$ such that $r!=\sqrt{2\pi r}\left(\frac re\right)^r e^{\frac \theta{12r}}$). Next, $\frac {e(a+b)}{r}\le \frac {3(a+b)}{r}\le c$. Next, $(2c)^r-c^r=c^r(2^r-1)\ge c^r$. At last, $c^r\ge a$, because $c\ge a^{\frac 1r}$.