Let $r<-1$ be a fixed real number and define the function $f:[1, +\infty)\rightarrow\mathbb{R}$ by $f(x) = -x^r$. In this case, $f$ is concave and increasing on $[1, +\infty)$. Also, we know $\zeta(r) = \sum_{k=1}^\infty \frac{1}{k^r}$ for $r\in \mathbb{C}$ and $\zeta(r, q) = \sum_{k=0}^\infty \frac{1}{(q+k)^r}$ for each fix complex $q$ with $Re(q)>0$. Now, my question is:
Are the following inequality established?
Case 1. \begin{align*} \zeta(-r, x) - \zeta(-r) + \frac{x^{r+1}}{r+1} \geq \frac{1}{r+1}, \end{align*} for $x\geq 1$.
Case 2. \begin{align*} \zeta(-r, x+1) - \zeta(-r)\geq \frac{-1-x^r}{2} - \int_{1}^x t^r dt, \end{align*} for $x\geq 1$.