求证一数学题

己知ABC=1,求证:1/(AB+A+1)+1/(BC+B+1)+1/(AC+C+1)=1

1/(AB+A+1)=1/(AB+A+ABC)=1/[A(B+1+BC)]

左边=1/(AB+A+1)+1/(BC+B+1)+1/(AC+C+1)

=1/[A(B+1+BC)]+1/(BC+B+1)+1/(AC+C+1)

=(1+A)/[A(BC+B+1)]+1/(AC+C+1)

=(1+A)/[A(BC+B+ABC)]+1/(AC+C+1)

=(1+A)/[AB(AC+C+1)]+1/(AC+C+1)

=(1+A+AB)/[AB(AC+C+1)]

=(ABC+A+AB)/[AB(AC+C+1)]

=A(BC+B+1)/[AB(AC+C+1)]

=(BC+B+1)/[B(AC+C+1)]

=(BC+B+ABC)/[B(AC+C+1)]

=B(AC+C+1)/[B(AC+C+1)]

=1
右边=1,左边=右边

所以原命题成立.证毕

1/ab+a+1 + 1/bc+b+1 + 1/ac+c+1
=abc/(ab+a+abc)+ 1/(bc+b+1) + 1/(ac+c+1)
=bc/(bc+b+1)+ 1/(bc+b+1) + 1/(ac+c+1)
=(bc+1)/(bc+b+1)+ 1/(ac+c+1)
=(bc+1)/(bc+b+abc)+ 1/(ac+c+1)
=(bc+1)/[b(c+1+ac)]+ b/[b(ac+c+1)]
=(bc+1+b)/[b(c+1+ac)]
=(bc+abc+b)/(bc+b+abc)
=1
或者
a/(ab+a+1)+b/(bc+b+1)+c/(ca+c+1)
=a/(ab+