设abc=1 求证(a/ab+a+1)+(b/bc+b+1)+(1/ac+c+1)=1

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设abc=1 求证(a/ab+a+1)+(b/bc+b+1)+(1/ac+c+1)=1
设abc=1 求证(a/ab+a+1)+(b/bc+b+1)+(1/ac+c+1)=1

设abc=1 求证(a/ab+a+1)+(b/bc+b+1)+(1/ac+c+1)=1
[a/(ab+a+1)]+[b/(bc+b+1)]+[c/(ac+c+1)] =[a/(ab+a+abc)]+[b/(bc+b+1)]+[c/(ac+c+1)] =[(1+b)/(bc+b+1)]+[1/(ac+c+1)] =[(1+b)abc/(bc+b+abc)]+[c/(ac+c+1)] =(ac+abc+c)/(ac+c+1) =(ac+1+c)/(ac+c+1) =1