证明:隐陵芦∵灶带abc=1∴1/ab+a+1 +1/bc+b+1 +1/ca+c+1=c/c(ab+a+1)+ ac/ac(bc+b+1) +1/(ca+c+1)=c/(abc+ac+c)+ ac/(abc²+abc+ac) +1/汪培(ca+c+1)=c/(ca+c+1)+ ac/ (ca+c+1)+ 1/(ca+c+1)=(ca+c+1)/(ca+c+1)=1