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.

As an alternative, by induction, after the base case, we can proceed as follows for the induction step assuming true that $\binom{n+m}{k}+\binom{n-m}{k}\ge 2\binom{n}{k}$, then
$$\binom{n+m+1}{k}+\binom{n-m-1}{k}=\binom{n+m}{k}+\binom{n+m}{k-1}+\binom{n-m}{k}-\binom{n-m-1}{k-1}\ge $$
$$\ge 2\binom{n}{k}+\binom{n+m}{k-1}-\binom{n-m-1}{k-1}\ge 2\binom{n}{k}$$
since $\binom{n+m}{k-1}-\binom{n-m-1}{k-1}\ge0$

Edit
This answer is related to the following deleted question from the user Andrew.
