Consider the inequality below
\begin{gather*} \frac{1}{2^{n+1}}\left[\binom{n+1}{k-1} + \binom{n+1}{k+1}\right] > \frac{1}{2^n}\binom{n}{k} \end{gather*}
It can be reduced through the following passages
\begin{gather*} \frac{1}{2}\left[\binom{n+1}{k-1} + \binom{n+1}{k+1}\right] > \binom{n}{k} \\ \frac{1}{2}\left[\binom{n+1}{k-1} + \binom{n}{k} + \binom{n}{k+1}\right] > \binom{n}{k}\\ \binom{n+1}{k-1} + \binom{n}{k+1} > \binom{n}{k} \\ \binom{n+1}{k-1} > \binom{n}{k} - \binom{n}{k+1} \end{gather*}
The last is trivially true for $k\leq \lfloor{\frac{n-1}{2}}\rfloor$, might you conclude the proof for the remaining values?
The context motivating the inequality
Doing an exercise about the Rademacher Processes $\{S_n\}$ I've asked to shown that $\mathcal{F}'_n=\sigma \left<S_n\right>$ is not a filtration. Considering that $\frac{S_n - n}{2}$ has a binomial distribution $\text{Bin}(n, \frac{1}2)$, $S_n$ is a simple function and the $\sigma$-algebra generated from it is the the $\sigma$-algebra generated by the partition of $[0, 1)$ given by the preimages of the singletons in its image. Now two $\sigma$-algebra generated by finite partitions are equal if and only if the partitions are equal, morover one is contained in the other only if the latter's partition is finer than the former's one. It's easy to show that for each $x$ in the support of $S_n$, $S_n^{-1}(x)\cap S_{n+1}^{-1}(x+1) \neq \emptyset$ and $S_n^{-1}(x)\cap S_{n+1}^{-1}(x-1) \neq \emptyset$, that means that a necessary condition to have the mentioned property on the partitions is that we must have $S_{n+1}^{-1}(x+1) \cup S_{n+1}^{-1}(x-1) \subseteq S_n^{-1}(x)$. I've tried to falsify that with an argument based on measure, the inequality at the beginnig, in that context means, $\mathbb{P}(S_{n+1}^{-1}(x+1) \cup S_{n+1}^{-1}(x-1))>\mathbb{P}(S_n^{-1}(x))$. I've shown that there are some values for wich it's true, i.e. some non-trivial set in $\sigma \left<S_n\right>$ not measurable in $\sigma \left<S_{n+1}\right>$, and it's enough to finish the exercise. Concluding the proof will show that every non trivial set in $\sigma \left<S_n\right>$ is not measurable in $\sigma \left<S_{n+1}\right>$ and viceversa.
You used $\dbinom n k \le \dbinom n{k+1}$ for $k\le \left\lfloor\dfrac {n-1}2\right\rfloor$.
Similarly for $k \ge \left\lfloor\dfrac {n+1}2\right\rfloor$ we have $\dbinom n k > \dbinom n{k+1}$.
That is, for $k \ge \left\lfloor\dfrac {n+1}2\right\rfloor + 1$:
$$\binom {n+1}{k-1} = \binom {n}{k-2} + \binom {n}{k-1} \ge \binom {n}{k-1} > \binom nk \ge \binom nk - \binom n{k+1}$$
This leaves us with the case $k = \left\lfloor\dfrac {n+1}2\right\rfloor$, where we aim to prove:
$$\binom {n}{k-2} + \binom {n}{k-1} + \binom {n}{k+1} > \binom nk$$
For $n$ odd, write $n = 2m-1$, $k = m$. Then: $$\binom {2m-1}{m-2} + \binom {2m-1}{m-1} + \binom {2m-1}{m+1} = \binom {2m-1}{m-2} + \binom {2m-1}{m} + \binom {2m-1}{m+1} > \binom{2m-1}m$$
as long as $0 \le m-2 < m+1 \le 2m-1$, that is, $m \ge 2$. This gives the edge case $n=k=1$.
For $n$ even, write $n = 2m$, $k = m$. Then:
$$\begin{align}\binom {2m}{m-2} + \binom {2m}{m-1} + \binom {2m}{m+1} &= \frac{(2m)!}{(m+2)!(m-2)!} + \frac{2(2m)!}{(m+1)!(m-1)!} \\&=\frac{(2m)!}{(m+2)!(m-1)!}(m-1+2(m+2)) \\&=\frac m{(m+1)(m+2)}\frac{(2m)!}{(m!)^2}(3m+1) \\&=\frac {m(3m+1)}{(m+1)(m+2)}\binom {2m}m\end{align}$$
Our inequality holds when $m(3m+1) > (m+1)(m+2)$, that is, when $2m^2-2m-2>0$.
Therefore it holds when $m > 1$, giving the edge case $n=2, k=1$.
Hopefully these edge cases are trivial to deal with in your proof.