∵b+c＞a,即a＜b+c,∴√（a-b-c)²=b+c-a
2√a²+b²+c²-2ab-2bc+2ac=2√(a-b+c)²,又∵a+c＞b∴2√a²+b²+c²-2ab-2bc+2ac=2√(a-b+c)²=2a-2b+2c,∵a+b＞c∴3|a+b-c|=3a+3b-3c,所以,原式=(b+c-a)-(2a-2b+2c)+(3a+3b-3c)=6b-4c