a/(a+m)+b/(b+m)-c/(c+m)(相減通分)
=[a(b+m)(c+m)+b(a+m)(c+m)-c(a+m)(b+m)]/[(a+m)(b+m)(c+m)]
因?yàn)槿切蜛BC三邊長是a ,b,c>0,且m為正數(shù)
所以分母[(a+m)(b+m)(c+m)]>0
又因?yàn)閍(b+m)(c+m)+b(a+m)(c+m)-c(a+m)(b+m)
=abc+abm+acm+am^2+abc+bam+bcm+bm^2-abc-cam-cbm-cm^2
=abc+(abm+bam)+(am^2+bm^2-cm^2)
因?yàn)閍+b>c(三角形兩邊之和大于第三邊)
所以am^2+bm^2=(a+b)m^2>cm^2
所以(am^2+bm^2-cm^2)>0
abc+(abm+bam)>0
所以a/(a+m)+b/(b+m)>c/(c+m)