Inequality $\binom{n+m}{k}+\binom{n-m}{k}\ge 2\binom{n}{k}$ Is it true that $\binom{n+m}{k}+\binom{n-m}{k}\ge 2\binom{n}{k}$?
I've been checking this for many cases in the Pascal triangle ($n,m,k\in\mathbb{N}^*$ such that $n-m\ge k$) but cannot prove formally.

Claim. The inequality
$$\binom{n+m}k+\binom{n-m}k \ge 2 \binom nk$$
holds for any integers such that $0\le m,k \le n$.
Proof. It is clear that this is true for $k=0$. So we will assume from now on that $k\ge1$.
Let us denote $$a_j=\binom{n+j}k+\binom{n-j}k$$ for $j=0,1,\dots,n$. We have $a_0=2\binom nk$. It suffices to show that the sequence $a_j$ is non-decreasing.
For this, we just compute
\begin{align*}
a_{j+1}-a_j&=\binom{n+j+1}k-\binom{n+j}k-\binom{n-j}k+\binom{n-j-1}k\\
&=\binom{n+j}{k-1}-\binom{n-j-1}{k-1} \ge 0.
\end{align*}
So we get $a_{j+1}-a_j\ge0$ and thus $a_{j+1}\ge a_j$ whenever $j \le n-1$ (and $k-1 \ge 0$.

You can find other approaches to this problem (or generalizations) here: 


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