取a=b=c=1，则a+1b＝b+1c＝c+1a＝2．所以①②不正确；取a=b=c=2，则a+1b，b+1c，c+1a均大于2．所以③不正确；由a+1b+b+1c+c+1a=(a+1a)+(b+1b)+(c+1c)≥2a•1a+2b•1b+2c•1c＝6．所以a+1b，b+1c，c+1a三个正数中至少...