<s>[INST] Baby Rudin: Theorem 3.20 (c) and (d) In theorem 3.20, Rudin offers a proof to the limits of the following sequences:
$$u_n = n^{\frac{1}{n}}$$
$$v_n = \frac{n^{\alpha }}{(1+p)^n}$$ with $\alpha$ a real number and $p > 0$.
In both the proofs, Rudin uses inequalities which I am not familiar with. For $u_n$, the following inequality is used:
$$(1+x)^n \ge \frac{n(n-1)}{2}x^2$$
For $v_n$:
$${n \choose k}p^k > \frac{n^kp^k}{2^kk!}\quad \quad(n>2k) $$
I have painstakingly come up with proofs for both of those inequalities, however, and this is especially true for the second inequality, my proofs are not very elegant nor do they give me any idea of the inequality came to be.
Surely Rudin doesn't pull these inequalities out of thin air. Where can I find a more comprehensive derivation of these inequalities? Where do they come from exactly (which field)?

Update:
Here are my proofs. They both rely on induction.
For the first one:
Suppose $x\ge0$. Then, for $n = 2$:
$$ (1+x)^2 = 1+2x+x^2 $$
$$\frac{2(2-1)}{2} x^2 = x^2$$
Since $x\ge0$, we have $(1+x)^n \ge \frac{n(n-1)}{2}x^2 $ for n = 2.
Suppose our inequality is true for a certain n. Then:
$$  (1+x)^n + nx^2 \ge \frac{n(n+1)}{2}x^2$$
Let $D = (1+x)^{n+1} - (1+x)^n - nx^2 $. Since $\forall n \in \mathbb{N}, \forall m = 0,1, ... n, {n+1 \choose m} \ge {n \choose m} $, and ${n+1 \choose 2} - {n \choose 2} - n = 0$, we have:
$$D \ge 0 \Rightarrow (1+x)^{n+1} \ge (1+x)^n + nx^2$$
So:
$$(1+x)^{n+1} \ge \frac{n(n+1)}{2}x^2$$ The theorem follows by induction.
For the second inequality:
We want to prove that $n(n-1)...(n-k+1) > \frac{n^k}{2^k}$, with $n > 2k$
Let $n\in\mathbb{N}, k$ and integer greater than 0 such that $2n > k$. Then for $k=1$, we have: $$n > \frac{n}{2} $$ 
Now suppose our inequality holds for a certain $k$. Then: 
$$n(n-1)...(n-k+1)(n-k) > \frac{n^k}{2^k}(n-k)$$
Since $n > 2k$, $n-k > \frac{n}{2} > 0$ such that:
$$n(n-1)...(n-k+1)(n-k) > \frac{n^k}{2^k}(n-k) > \frac{n^{k+1}}{2^{k+1}} $$
By induction, our inequality holds for all $k$ such that $n > 2k$
 [/INST] For the first one, assuming $x\geq0$ :
$$
\begin{align}
(1+x)^n &= \sum_{k=0}^n \binom{n}{k}x^k \\
&=1+nx+\frac{n(n-1)}{2}x^2+\cdots+x^n \\
&\geq \frac{n(n-1)}{2}x^2
\end{align}
$$
For the second one :
$$
\begin{align}
\binom{n}{k}&= \frac{n(n-1)\cdots(n-k+1)}{k!} \\
&=\frac{1\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)}{k!}n^k \\
&>\frac{(1/2)(1/2)\cdots(1/2)}{k!}n^k \\
&=\frac{n^k}{2^kk!}
\end{align}
$$
</s>