Show that $f_{d}(n+1)\ge f_{d}(n)$ Let $n,d$ are non-negative Integers with $n>2$
$$f_{d}(n)=\Bigg(n-\Bigg\lceil\frac{\sum_{i=1}^n i^d}{n^d}\Bigg\rceil\Bigg) $$


Show that
    $$
\Bigg(n+1 -\Bigg\lceil\frac{\sum_{i=1}^{n+1} i^d}{(n+1)^d}\Bigg\rceil\Bigg) - \Bigg(n-\Bigg\lceil\frac{\sum_{i=1}^n i^d}{n^d}\Bigg\rceil\Bigg) \geq 0.
$$
is it equivalent to show
    $$
\Bigg(n+1 - \frac{\sum_{i=1}^{n+1} i^d}{(n+1)^d}\Bigg) - \Bigg(n-\frac{\sum_{i=1}^n i^d}{n^d}\Bigg) \geq 0\ \ \ ?
$$


This is true since
$$
\frac{\sum_{i=1}^{n+1} i^d}{(n+1)^d} = \frac{(n+1)^d + \sum_{i=1}^n i^d}{(n+1)^d} = 1 + \frac{\sum_{i=1}^n i^d}{(n+1)^d} < 1 + \frac{\sum_{i=1}^n i^d}{n^d}.
$$
Main problem here
 A: First, note
$$a \ge b \implies \lceil a \rceil \ge \lceil b \rceil \tag{1}\label{eq1}$$
To see this, let $a = n_1 + r_1$ and $b = n_2 + r_2$ where $n_1, n_2 \in \mathbb{Z}$, $n_1 \ge n_2$, $r_1, r_2 \in \mathbb{R}$ and $0 \le r_1,r_2 \lt 1$. If $n_1 \gt n_2$, then \eqref{eq1} is obviously true. If $n_1 = n_2$, then $r_1 \ge r_2$. If $r_1 = 0$, then $r_2 = 0$, so $\lceil a \rceil = n_1 = n_2 = \lceil b \rceil$. If $r_1 \gt 0$, then $\lceil a \rceil = n_1 + 1 = n_2 + 1 \ge \lceil b \rceil$.
Also, $\lceil n + c \rceil = n + \lceil c \rceil$ for all $n \in \mathbb{Z}$ and $c \in \mathbb{R}$.
Using this, from what you've already shown,
$$\begin{equation}\begin{aligned}
1 + \frac{\sum_{i=1}^n i^d}{n^d} & \gt \frac{\sum_{i=1}^{n+1} i^d}{(n+1)^d} \\
\left\lceil 1 + \frac{\sum_{i=1}^n i^d}{n^d} \right\rceil & \ge \left\lceil \frac{\sum_{i=1}^{n+1} i^d}{(n+1)^d} \right\rceil \\
1 + \left\lceil \frac{\sum_{i=1}^n i^d}{n^d} \right\rceil & \ge \left\lceil \frac{\sum_{i=1}^{n+1} i^d}{(n+1)^d} \right\rceil \\
1 - \left\lceil \frac{\sum_{i=1}^{n+1} i^d}{(n+1)^d} \right\rceil + \left\lceil \frac{\sum_{i=1}^n i^d}{n^d} \right\rceil & \ge 0 \\
n + 1 - \left\lceil \frac{\sum_{i=1}^{n+1} i^d}{(n+1)^d} \right\rceil - n + \left\lceil \frac{\sum_{i=1}^n i^d}{n^d} \right\rceil & \ge 0 \\
\left(n + 1 - \left\lceil \frac{\sum_{i=1}^{n+1} i^d}{(n+1)^d} \right\rceil\right) - \left(n - \left\lceil \frac{\sum_{i=1}^n i^d}{n^d} \right\rceil\right) & \ge 0
\end{aligned}\end{equation}\tag{2}\label{eq2}$$
Thus, the second part being true shows that the first part you're asking about is also true.
