I found the interesting inequality when I study hypergraph 2-coloring
$$\sum\_{i+j=k} \binom{r-1}{i}\binom{r-1}{j}(1-p)^i(1+p)^j \leq \binom{2r-2}{k}$$


$0\leq i, j < r$, $0\leq p \leq 1$. I want to know how to proof it.



Combined identity perturbation
The left-hand side is the coefficient of $x^k$ in
$$
\left(1+(1-p)x\right)^{r-1}\left(1+(1+p)x\right)^{r-1}=\left(1+2x+(1-p^2)x^2\right)^{r-1}\ .
$$
This coefficient can be obtained via the Multinomial Theorem as
$$
\sum\_{a,b}\mathbf{1}\left\{
\begin{array}{c}
a,b\ge 0 \\
a+b\le r-1 \\
a+2b=k
\end{array}
\right\}\frac{(r-1)!}{(r-1-a-b)!a!b!} 2^a (1-p^2)^b
$$
where $\mathbf{1}\{\cdots\}$ is the indicator function of the conditions between the braces. Now it is immediate that the left-hand side is a decreasing function of $p$ on the interval $[0,1]$. It is thus bounded above by the value at $p=0$ which is equal to the right-hand side, by the Chu-Vandermonde identity.


