$\sum\limits_{i=1}^{2019}|a_i|-|\sum\limits_{i=1}^{2019}a_i|$的取值范围

hbghlyj

(1)实数$a_i(i=1,2\cdots2019)$满足$|a_i|\leq1$,求$\sum\limits_{i=1}^{2019}|a_i|-|\sum\limits_{i=1}^{2019}a_i|$的取值范围
(2)复数$a_i(i=1,2\cdots2019)$满足$|a_i|\leq1$,求$\sum\limits_{i=1}^{2019}|a_i|-|\sum\limits_{i=1}^{2019}a_i|$的取值范围

hbghlyj

(1)设$P=\sum\limits_{a_i>0\\1\le i\le2019}a_i,N=\sum\limits_{a_i<0\\1\le i\le2019}a_i,S=\sum\limits_{i=1}^{2019}|a_i|-|\sum\limits_{i=1}^{2019}a_i|=P-N-|P+N|$,若P+N$\ge$0,则$S=-2N\in[0,2018]$,当且仅当1010个$\in[0,1]$且和不小于1009,1009个-1,或者,1009个1,1010个$\in[-1,0]$且和为-1009时取最大值,当且仅当全$\in[0,1]$时取最小值.若P+N<0,则S=2P$\in[0,2018]$,当且仅当1010个$\in[0,1]$且和为1009,1009个-1,或者,1009个1,1010个$\in[-1,0]$且和不大于-1009时取最大值,当且仅当全$\in[-1,0]$时取最小值.
请看，这个论证有问题吗？{:blunder:}

hbghlyj

回复 2# hbghlyj
(2)$\sum\limits_{i=1}^{2019}|a_i|-|\sum\limits_{i=1}^{2019}a_i|\le\sum\limits_{i=1}^{2019}|a_i|\le2019$,当$a_i=e^{\frac {2πi}{2019}},i=1,2,\cdots,2019$时取等
所以$\sum\limits_{i=1}^{2019}|a_i|-|\sum\limits_{i=1}^{2019}a_i|\in[0,2019]$