Question

已知实数a b c,满足a+b+c=0,a>b>c,求证1＼3<a\(a-c)<2\3

Answer 1

由题可得，a＞0，c＜0
a＋c＝－b。
由a＞b＞c可得－a＜－b＜－c；所以－a＜a＋c＜－c
可得1／2＜－c／a＜2
可得3／2＜1－c／a＜3
即3／2＜(a－c)／a＜3
利用倒数可得1／3＜a／(a－c)＜2／3

Answer 2

a+b+c>0 & a>b>c
=>2a+c>0
=>3a>a-c
a>c => a/(a-c)>1/3

a+b+c>0 & a>b>c
=>a+2c>0
=>3a<2a-2c
=>a/(a-c)<2/3