Estimate of an integral in one dimension

Question

Let $s,\sigma,t\in\mathbb{R}$, with $s<\sigma<t$. I want to prove the following estimate $$\int_s^t (t-\sigma)^{N+1}(\sigma-s)^{N+1}d\sigma\leq \frac{\Gamma(N+2)^2}{\Gamma(2(N+2))}\vert t-s\vert^{2(N+2)-1},$$ where $N\in\mathbb{N}$, and $\Gamma$ is the Gamma Function. Any suggestion to obtain the constant $\frac{\Gamma(N+2)^2}{\Gamma(2(N+2))}$?

Answer 1

The following does not work (there is a reverse inequality), but maybe can be useful? (the answer is in the comments below)

Notice that $$ \frac{\Gamma(2N+4)}{\Gamma^2(N+2)} = \frac{(2N+3)!}{(N+1)!^2} = (2N+3) {2N+2 \choose N+1} \ge \sum_{k=0}^{2N+2} {2N+2 \choose k} = 2^{2N+2}. $$ Also, by the substitution $\sigma = (t-s) x + s$ one finds $$ I=\int_s^t (t-\sigma)^{N+1}(\sigma -s)^{N+1} \, d\sigma = (t-s)^{2N+3}\int_0^1 (1-x)^{N+1} x^{N+1}\, dx. $$ Now notice that $(1-x)x\le 1/4$ so that $$ I\le (t-s)^{2N+3} \frac{1}{4^{N+1}} = \frac{(t-s)^{2N+3}}{2^{2N+2}}. $$