设a=y+z,b=x+z,c=x+y(x,y,z>0)=>x+z+2yz/2时 （x+z)/2>=2x-z =>x=y>=2x-z当00b/a=(x+z)/(y+z)>=(x+z)/((x+z)/2+z) =(2x+2z)/(x+3z)=2-4z/(x+3z)=2-4/((x/z)+3)显然,x02-4/3=2/3当00,b/a=(x+z)/(y+z)=2x =>y/z+1>=2x/...