设x,y,z是三角形ABC的三边,化简Ix+y-zI+Ix-y+zI+Ix-y-zI.II为绝对值符号

Question

设x,y,z是三角形ABC的三边,化简Ix+y-zI+Ix-y+zI+Ix-y-zI.
I I为绝对值符号

Answer 1

三角形两边之和大于第三边
所以x+y-z>0,x+z-y>0,x-(y+z)<0
所以Ix+y-zI+Ix-y+zI+Ix-y-zI
=(x+y-z)+(x-y+z)-(x-y-z)
=x+y+z