I am would like to prove ${n \choose k} \le {n \choose k-1} +{n \choose k+1}$.
I can use the monotonicity property of the binomial coefficient to show this.
However, I wonder if there is an easy way to see this "immediately".
(Perhaps a combinatorial proof?)
Any suggestions are welcome.
Given a $k$-subset $S \subset \{1,\ldots, n\}$, define $f(S)$ as follows: if $1 \in S$, let $f(S) = S - \{1\}$, otherwise, let $f(S) = S \cup \{1\}$. Observe that $f$ is an injection and always $|f(S)| \in \{k-1, k+1\}$.
This is based on Did's comment:
By an easy (and very well-known) combinatorial argument, we have $$ {n\choose k}={n-1\choose k}+{n-1\choose k-1}.$$
The following two inequalities are also immediate due to combinatorial reasons:
$${n-1\choose k} \le {n\choose k+1}.$$
(To any choice of $k$ out of $n-1$, append the extra ball).
$${n-1\choose k-1} \le {n\choose k-1}.$$
(Just don't choose the last ball).
Putting these two together, we get $$ {n\choose k} \le {n\choose k+1}+{n\choose k-1}$$.
Directly: $$C_k^n\le C_{k-1}^n+C_{k+1}^n=\underbrace{C_{k-2}^{n-1}}_{0 \ if \ k-2<0}+\underbrace{C_{k-1}^{n-1} +C_{k}^{n-1}}_{C_k^n}+\underbrace{C_{k+1}^{n-1}}_{0 \ if \ k+1>n}.$$
