Rearrangement inequality for sum

Question

Rearrangement inequality: Assume we have finite ordered sequences of nonnegative real numbers $0 \le a_1 \le a_2 \le\cdots\le a_n \quad\text{and}\quad 0\le b_1 \le b_2 \le\cdots\le b_n, \cdots\,, \quad\text{and}\quad 0\le z_1 \le z_2 \le\cdots\le z_n$ and a permutation $b_{\sigma(1)}, b_{\sigma(2)}, \ldots,b_{\sigma(n)}$ of $b_1, b_2\dots,b_n$,$\cdots$ , and another permutation $z_{\tau(1)}, z_{\tau(2)}, \dots,z_{\tau(n)}$ of $z_1, z_2, \dots,z_n$, then $$ a_1 b_{\sigma(1)}\cdots z_{\tau(1)} + \cdots + a_n b_{\sigma(n)} \cdots z_{\tau(n)} \le a_1 b_1 \cdots z_1 + \cdots + a_n b_n \cdots z_n.$$

Does this inequality hold if we replace multiplication by sumation? detail as follows:

Assume we have finite ordered sequences of real numbers $ a_1 \le a_2 \le\cdots\le a_n \quad\text{and}\quad b_1 \le b_2 \le\cdots\le b_n, \cdots\,, \quad\text{and}\quad z_1 \le z_2 \le\cdots\le z_n$ and a permutation $b_{\sigma(1)}, b_{\sigma(2)}, \ldots,b_{\sigma(n)}$ of $b_1, b_2\dots,b_n$,$\cdots$ , and another permutation $z_{\tau(1)}, z_{\tau(2)}, \dots,z_{\tau(n)}$ of $z_1, z_2, \dots,z_n$, then

$$ |a_1 + b_{\sigma(1)}+\cdots +z_{\tau(1)}| + \cdots + |a_n + b_{\sigma(n)}+ \cdots + z_{\tau(n)}| \le |a_1 + b_1 + \cdots + z_1| + \cdots + |a_n + b_n + \cdots + z_n|$$

Answer 1

Yes, and moreover, for arbitrary convex function $f$ defined on the interval between $a_1+\dots+z_1$ and $a_n+\dots+z_n$ we have $$f(a_1 + b_{\sigma(1)}+\cdots +z_{\tau(1)}) + \cdots + f(a_n + b_{\sigma(n)}+ \cdots + z_{\tau(n)}) \leqslant f(a_1 + b_1 + \cdots + z_1) + \cdots + f(a_n + b_n + \cdots + z_n).$$ This follows immediately from Karamata's inequality: for every $k=1,\dots,n$ the sum of any $k$ arguments of $f$ in LHS does not exceed the sum of the last $k$ arguments of $f$ in RHS, and for $k=n$ these sums are equal.