Upper bound for $\sum_{n=1}^xn^{k-1}$ From some Calculus and guess-work, I found that
$$k\sum_{n=1}^xn^{k-1}<(x+\frac12)^k\tag1$$
In fact, I found that it was very, very, close.
And, from even more guesswork,
$$\lfloor(x+\frac12)^k\rfloor-1\le k\sum_{n=1}^xn^{k-1}\le\lfloor(x+\frac12)^k\rfloor\tag2$$
I want to know if what I have is actually correct, because if so, then this is a great bound to the summation and is off by $\pm1$, which I find the summation for large $k$ has $\lfloor(x+\frac12)^k\rfloor-1=k\sum_{n=1}^xn^{k-1}$, but I have no way of proving this.
EDIT
$k\in\mathbb{R}$
It appears that $(2)$ becomes false for decently large $k,x$, as Slade has shown.  It fails at $k=3,x=8$.
However, I would like to know if $(1)$ is still provable and whether or not it works.
Also, for what values of $k$ does $(2)$ start to fail?  I can see from my calculator that it starts to fail around $k=2.4$, which I managed to input $x=1000$, but to see if a number like $k=2.3$ fails would require my calculator to manage some very large numbers... and that's not quite a proof.
I can also see that it doesn't fail for $k=2$ because of Faulhaber's formula, but I'm not quite sure about $k=2.3$ or something.
 A: Since $t^{k-1}$ is convex for $k\not\in(1,2)$, Jensen's Inequality says
$$
\begin{align}
\left(n+\tfrac12\right)^k-\left(n-\tfrac12\right)^k
&=\int_{n-\frac12}^{n+\frac12}kt^{k-1}\,\mathrm{d}t\\[3pt]
&\ge kn^{k-1}\tag{1}
\end{align}
$$
Summing $(1)$ yields
$$
\begin{align}
\sum_{n=1}^xkn^{k-1}
&\le\left(x+\tfrac12\right)^k-\left(\tfrac12\right)^k\\
&\lt\left(x+\tfrac12\right)^k\tag{2}
\end{align}
$$
Thus, the first inequality is true except possibly for $1\lt k\lt2$.

Offset Euler-Maclaurin
There is an offset version of the Euler-Maclaurin Sum Formula that says
$$
\begin{align}
\sum_{k=1}^nf(k)
&\sim\left.\int f(x)\,\mathrm{d}x\,\right|_{n+\frac12}+C-\tfrac1{24}f'\!\left(n+\tfrac12\right)+\tfrac7{5760}f'''\!\left(n+\tfrac12\right)\\
&-\tfrac{31}{967680}f^{(5)}\!\left(n+\tfrac12\right)+\tfrac{127}{154828800}f^{(7)}\!\left(n+\tfrac12\right)-\tfrac{73}{3503554560}f^{(9)}\!\left(n+\tfrac12\right)\tag{3}
\end{align}
$$
When $f(n)=kn^{k-1}$, it can be shown that $C=k\zeta(1-k)$; therefore,
$$
\sum_{n=1}^xkn^{k-1}\sim\left(x+\tfrac12\right)^k+k\zeta(1-k)-\frac1{12}\binom{k}{2}\left(x+\tfrac12\right)^{k-2}+\frac7{240}\binom{k}{4}\left(x+\tfrac12\right)^{k-4}\tag{4}
$$
Asymptotic Expansion $(4)$ shows that
$$
\lim_{x\to\infty}\left[\sum_{n=1}^xkn^{k-1}-\left(x+\tfrac12\right)^k\right]=-\infty
$$
for $k\gt2$. Therefore, the lower bound in the question cannot hold for any constant.
A: Your second equation is definitely false for $k\geq 3$, though it is probably a good estimate for $k=1,2$.
Note that when $k=1$, $k\sum_{n=1}^x n^{k-1} = x$, while $(x+1/2)^k =x+1/2$
Also, when $k=2$, $k\sum_{n=1}^x n^{k-1} = x(x+1)$, while $(x+1/2)^k =x^2 + x + 1/4$.  So in both these cases, the expressions differ by a constant.
However, when $k=3$, $k\sum_{n=1}^x n^{k-1} = x^3 + \frac{3}{2}x^2 + \frac{1}{2}x$, while $(x+1/2)^k = x^3 + \frac{3}{2}x^2 + \frac{3}{4}x + \frac{1}{8}$.  So in this case $(x+1/2)^k$ will grow faster, and eventually the inequality will be false. (In general, we can use Faulhaber's formula to show that this problem always occurs for $k\geq 3$, because $B_2 = \frac{1}{6} < \frac{1}{2^2}$.)
For example, if we set $k=3$, $x=8$, we have $\lfloor (x+1/2)^k \rfloor - 1 = 613$, while $k\sum_{n=1}^x n^{k-1} = 612$.
A: Let's consider the statement
$$
k\sum_{n=1}^xn^{k-1}\leq (x+\frac12)^k\tag1
$$
with $x \in \mathbb N$ and $k\geq 2$, $k \in \mathbb R$.
We prove it by induction on $x$.


*

*Induction base: for $x=1$ the statement becomes:
$$
k\cdot 1^{k-1}\leq (1+\frac12)^k
$$
that is true for any $k \geq 2$

*Inductive step: we assume that (1) is true for some $x$ and $k \geq 2$ and show that it must hold also for $x+1$ and $k \geq 2$.
This lemma will help:

If $f$ is convex then $\int_a^b f(t)\geq (b-a)f(\frac {a+b}2)$

Proof of the inductive step: first we have
$$
k\sum_{n=1}^{x+1}n^{k-1}-k\sum_{n=1}^xn^{k-1}=k(x+1)^{k-1}
$$
and on the other hand by the lemma
$$
(x+1+\frac12)^k-(x+\frac12)^k=\int_{1/2}^{3/2} k(x+t)^{k-1}dt\geq k(x+1)^{k-1}
$$
since for $k\geq 2$ the function $f:t \mapsto (x+t)^{k-1}$ is convex.

Proof of the lemma for $f$ derivable (relevant for our case):
Let $t_0=\frac{a+b}2$. $f$ convex and differentiable implies
$$
f(t)=f(t_0)+\int_{t_0}^t f'(u)du \geq f(t_0)+f'(t_0)(t-t_0)
$$
(that is $f$ is above the tangent line in $t_0$) therefore
$$
\int_a^b f(t)dt \geq \int_a^b [f(t_0)+f'(t_0)(t-t_0)]=(b-a)f(t_0)
$$
