German MO combinatorics problem 1995

Question

I found this old German MO problem in a book but no answer is given. Here is the problem : Prove that for all integers $k$ and $n$ with $1\le k\le 2n$ : $$\binom{2n+1}{k-1}+\binom{2n+1}{k+1}\ge 2\cdot \frac{n+1}{n+2}\cdot \binom{2n +1}{k}$$ Does anyone know how to solve it or where could i find the answer ?

Answer 1

Just to clarify user 10354138 answer beacause there might be a small mistake:

$$\frac{k}{2n-k+2}+\frac{2n-k+1}{k+1}\ge 2\cdot \frac{n+1}{n+2}$$ $$\iff \frac{k}{2n-k+2}+\frac{2n-k+1}{k+1}-\frac{2(n+1)}{n+2}\ge0$$ $$\iff \frac{k(k+1)(n+2)+(n+2)(2n-k+1)(2n-k+2)-2(n+1)(2n-k+ 2)(k+1)}{(2n-k+2)(k+1)(n+2)}\ge0$$

As from the statement $2n\ge k\ge1$, we know that the denominator is strictly positive. It remains to show that the numerator is positive: $$ k^2n+2k^2+kn+2k+(2n^2-kn+5n-2k+2)(2n-k+2)-(2nk+2n+2k+2)(2n-k+2) \ge0$$ $$\iff 4n^3+10n^2-8kn^2+6n-16kn+4k^2n+6k^2-6k\ge0$$ $$\iff 2(n-k)(2n+3)(n-k+1)\ge0$$

Answer 2

Dividing by $\binom{2n+1}{k}$, it suffices to prove $$ \frac{k}{2n-k+2}+\frac{2n-k+1}{k+1}\geq2\cdot\frac{n+1}{n+2}. $$

Transfer everything to LHS and putting everything under same denominators, this is equivalent to $$ \frac{2(2n+3)(n-k)(n+1-k)}{(k+1)(2n+2-k)(n+2)}\geq 0. $$ For integers $n,k$ satisfying $1\leq k\leq 2n$, the denominator is clearly positive, and the numerator is positive for $k\neq n,n+1$ where it is $0$, as desired.

Answer 3

If we add $2{2n+1\choose k}$ to both sides we get $${2n+3\choose k+1}\geq\frac{4n+6}{n+2}{2n+1\choose k}$$ and $$(n+1)(n+2)\ge(k+1)(2n-k+2)$$ and $$(n-k)^2\geq k-n$$ which is true for all integers.