Prove:$|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$ 
Prove:$|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$

example1: $|x-1|+|x-2|\geq 1$
my solution：(substitution)
$x-1=t,x-2=t-1,|t|+|t-1|\geq 1,|t-1|\geq 1-|t|,$
square,
$t^2-2t+1\geq 1-2|t|+t^2,\text{Since} -t\leq -|t|,$
so proved.
question1 : Is my proof right? Alternatives?
one reference answer: 
$1-|x-1|\leq |1-(x-1)|=|1-x+1|=|x-2|$
question2 : prove:
$|x-1|+|x-2|+|x-3|\geq 2$
So I guess:( I think there is a name about this, what's that? wiki item?)
$|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$
How to prove this? This is  question3. I doubt whether the two methods I used above may suit for this general case.
Of course, welcome any interesting answers and good comments.
 A: Note that the sum $F(x)=|x-1|+|x-2| +\cdots +|X-n|$ is the sum of the distances from $x$ to the points $1,2,\dots,n$. Draw the $n$ points $1,2,3,\dots,n$ on the number line, taking $n$ say $7$ or $8$. 
Imagine that a particle $P$  starts far to the left of $1$, and travels to the right.  
Until it hits $x=1$, the sum is of the distances of $P$ from $1,2,\dots, n$ is decreasing. At $x=1$ it becomes $1+2+\cdots+(n-1)$. 
As we travel from $1$ to $2$, the function $F(x)$ is decreasing. For each tiny step $s$ we take to the right increases our distance from  $1$ by $s$, but it decreases our distance from each of $2, 3,\dots,n$ by $s$. So each tiny step $s$ we take decreases $F(x)$ by $(n-1)s-s$.
If $n$ is not too small, this decrease continues. For each small step $s$ we take from $2$ towards $3$ increases our distance from $1$ and $2$ by $s$, and decreases our distance from each of the other points by $s$, for a decrease of $(n-2)s-2s$.
If $n$ is odd, $F(x)$ keeps decreasing until $x=\frac{n+1}{2}$, and then by symmetry $F(x)$ starts to increase. If $n$ is even, then $F(x)$ reaches a minimum at all points between $\frac{n}{2}$ and $\frac{n+1}{2}$.
For $n$ odd, say $n=2k+1$, the minimum value of $F(x)$ is $2(1+2+3+\cdots +k)$. This is $k(k+1)$. For $n$ even, say $n=2k$,  the minimum value of $F(x)$ is $(1+2+\cdots +(k-1))+(1+2+\cdots +k)$. This is $k^2$.
Back to the question in the post.  
We  want to show that if $n$ is odd, say $n=2k+1$, then $k(k+1) \ge 2k$, and that if $n$ is even, say $2k$m then $k^2\ge 2k-1$. Both of these are obvious.
Remark: We wrote out a solution in order to emphasize the geometry
of the situation.
Generalization: Suppose that instead of $1,2,\dots,n$ we have numbers $a_1 \le a_2\le a_3\le \cdots \le a_n$.
Play the same walking game. If $n=2k+1$ is odd, then $F(x)$ reaches its minimum at $x=k+1$. The number $a_{k+1}$ is the median of our $n$ numbers $a_1,\dots, a_n$. 
If $n$=2k is even, then $F(x)$ reaches a minimum at all points between $x=k$ and $x=k+1$. Any such $x$ can be viewed as a median of the $a_i$. 
A: After the answer of Thomas Andrews my answer appears to be dispensable. But I'll leave it here as a poor alternative. Here a proof valid for any $ x $ since $ n $ is even. For the case $ n $ odd simply discard the parcel | x-n |. 
\begin{align}
|x-1|+|x-2|+|x-3|+\ldots+|x-n|
=
&
|x-1|+|2-x|+|x-3|+
\\
+
&
\ldots+|(-1)^{n+1}x+(-1)^{n}n|
\\
\geq
&
\bigg[(1-x)+(x-2)+(3-x)+
\\
+
&
\ldots+(-1)^{n}x+(-1)^{n+1}n\bigg]
\\
=
&
(-1)^1+(-1)^{2}2+(-1)^{3}3+(-1)^{4}4+\ldots
\\
+
&
(-1)^{(n-1)}(n-1)+n(-1)^{n}
\\
=
&
\frac{n}{2}+n(-1)^{n}= \frac{n}{2}+n\geq n
\end{align}
A: (Edited)
Given a real-valued data set ${\bf y}=(y_k)_{1\leq k\leq n}$ the function
$$f(x):=\sum_{k=1}^n |x-y_k|$$
assumes its minimum at the median $\mu$ of ${\bf y}$. When $y_1\leq y_2\leq \ldots\leq y_n$ and $n=2m+1$ then $\mu=y_{m+1}$ (the "middle value"), and if $n=2m$ then $f$ is actually constant on the interval $[y_m,y_{m+1}]$, and one defines $\mu:=(y_m+y_{m+1})/2$.
Proof. When there are $r$ more $y_k$ to the right of $x$ than to the left of $x$ then increasing $x$ by $\Delta x$ will decrease $f$ by $r\cdot\Delta x$, and similarly, when there are $r$  more $y_k$ to the left of $x$ than to the right of $x$ then increasing $x$ by $\Delta x$ will increase $f$ by $r\cdot \Delta x$.$\qquad\square$
It follows that your sum $s(x)$ is minimal at $x={n+1\over2}$. It is then easily computed to $s_{\min}=n^2/4$ when $n$ is even and $s_{\min}=(n^2-1)/4$ when $n$ is odd.
When $n=1$ we have $s_{\min}=0=n-1$, and in the case $n=2$ we have $s_{\min}=1=n-1$. For $n\geq3$ we can argue as follows:
$$s_{\min}\geq{n^2-1\over 4}={(n-2)^2-1\over4}+n-1\geq n-1\ .$$
A: You want
$S(x) 
=\sum_{i=1}^n |x-i|$.
I will show
$\begin{cases}
\text{if } x \le 1&S(x) \ge \dfrac{n(n-1)}{2}\\
\text{if } x \ge n &S(x) \ge \dfrac{n(n-1)}{2}\\
\text{otherwise} &S(x) \ge \dfrac{n^2-1}{4}
\end{cases}
$
Exact results are
$\begin{cases}
\text{if } x \le 1, & S(x) = n(1-x) + \dfrac{n(n-1)}{2}\\
\text{if } x \ge n, & S(x) =  n(x-n)+ \dfrac{n(n-1)}{2}\\
\text{if } j\le x < j+1 \text{ where } 1 \le j < n, &S(x) =  \dfrac{n^2+(n-2x+1)^2-(1-2y)^2}{4}\text{ where } y=x-j
\end{cases}
$
As a check, these are correct for
$x=0, 1, n$, 
and
$n+1$.
We consider three cases:
$x \le 1$,
$x \ge n$,
and $1 < x < n$.
If $x \le 1$,
then
$|x-i| = -x+i$,
so
$\begin{align}
S(x) &= \sum_{i=1}^n (-x+i)\\
&=-nx + \frac{n(n+1)}{2}\\
&=n(1-x)-n+ \frac{n(n+1)}{2}\\
&=n(1-x) +\frac{n(n-1)}{2}\\
\end{align}
$
If $x \ge  n$,
then
$|x-i| = x-i$,
so
$\begin{align}
S(x) = \sum_{i=1}^n (x-i)
&=nx - \frac{n(n+1)}{2}\\
&=n(x-n)+n^2 - \frac{n(n+1)}{2}\\
&=n(x-n) \frac{2n^2-n(n+1)}{2}\\
&= n(x-n)+\frac{n^2-n}{2}\\
&= n(x-n)+\frac{n(n-1)}{2}\\
\end{align}
$
If $j \le x < j+1$
where $1 \le j < n$,
let $y = x-j$,
so $0 \le y < 1$.
Then
$\begin{align}
S(x)
&=\sum_{i=1}^n |x-i|\\
&=\sum_{i=1}^n |j+y-i|\\
&=\sum_{i=1}^j |j+y-i|+\sum_{i=j+1}^n |j+y-i|\\
&=\sum_{i=1}^j (j+y-i)+\sum_{i=j+1}^n -(j+y-i)\\
&=\sum_{i=1}^j (j+y-i)+\sum_{i=j+1}^n (i-j-y)\\
&=jy+\sum_{i=0}^{j-1} (i)-(n-j)y+\sum_{i=1}^{n-j} (i)\\
&=(2j-n)y+\frac{j(j-1)+(n-j)(n-j+1)}{2}\\
&=(2j-n)y+\frac{j^2-j+(n-j)^2+(n-j)}{2}\\
&=(2j-n)y+\frac{(n^2+(n-2j)^2)/2+(n-2j)}{2}\quad (*)\\
&=(2j-n)y+\frac{n^2+(n-2j)^2+2(n-2j)}{4}\\
&=\frac{n^2+(n-2j)^2+2(n-2j)-4y(n-2j)}{4}\\
&=\frac{n^2+(n-2j)^2+2(1-2y)(n-2j)}{4}\\
&=\frac{n^2+(n-2j+1-2y)^2-(1-2y)^2}{4}\\
&=\frac{n^2+(n-2x+1)^2-(1-2y)^2}{4}\\
&\ge \frac{n^2-1}{4}\\
\end{align}
$
Note: The line marked "(*)" used
$a^2+b^2 = \dfrac{(a+b)^2+(a-b)^2}{2}$.
A: Actually, $|x-n| + |x-1| = |n-x| + |x-1| \geq |n-1|=n-1$. Don't need the other terms.
Your first proof is correct. You have to take some care moving from $U^2\geq V^2$ to $U\geq V$, but in this case, $U=|t-1|$ is non-negative.
