My question is:
Show that for all $|x-1|+|x-2|+\dots+|x-10| > 23$
I have solved above problem as below,
If $x-1 > 0$ and $x-2 > 0$ and ......$x-10 > 0$ then
LHS $= x-1+x-2+x-3+\dots+x-10 = 10x-55 > 23$ (because $x>10$)
If $x-1<0$ and $x-2 < 0$ and ......$x-10 < 0$ then
LHS $= -x+1-x+2-x+3.....-x+10 = -10x+55 > 23 $ (because $x<1$)
If the above solution is wrong, please give me the correct method
On
Hint: by the triangle inequality $|x-1|+|x-10| = |x-1|+|10-x| \ge |x-1+10-x|=9$. The same way, $\,|x-2|+|x-9|\ge 7\,$, $\,|x-3|+|x-8|\ge 5\;\ldots$
On
The function is convex and $f({11 \over 2}+x) = f({11 \over 2}-x)$. Hence a $\min$ occurs at $x={11 \over 2}$, or $f({11\over 2}) = 25$.
(In fact, the $\min$ occurs for $x \in [5,6]$.)
On
The LHS is a piecewise linear function, so it can only achieve a minimum at a "turning" point, when one of the arguments is zero.
We have
$$|1-1|+|2-1|+\cdots|1-10|=0+1+\cdots 9=45,$$
$$|2-1|+|2-2|+\cdots|2-10|=1+0+1+\cdots 8=37,$$
$$\cdots$$
and the minimum is achieved with
$$|5-1|+|5-2|+\cdots|5-10|=4+3+2+1+0+1+\cdots 5=25.$$
Algebraically, we have
$$S_k=\sum_{i=1}^{k-1}(k-i)+\sum_{i={k+1}}^{10}(i-k)=k^2-11k+55$$
and the minimum is found on either sides of the continuous minimum.
Just use triangle inequality:
$$|x-1|+|x-2|+\dots+|x-10| = |1-x|+|2-x|+\dots+|5-x|+ |x-6|+|x-7|\dots+|x-10| >$$ $$|1-x+2-x+...+x-10| = |15-6-7-8-9-10| = |-25|$$