a除以(ab+a+1)+c除以(ac+c+1)+b除以(bc+b+1)(1)
=ac/(abc+ac+c)+cb/(abc+cb+b)+ba/(abc+ab+a)
=ac/(1+ac+c)+cb/(1+cb+b)+ba/(1+ab+a)(2)
=acb/(b+acb+cb)+acb/(a+acb+ab)+cba/(c+abc+ac)
=1/(b+1+cb)+1/(a+1+ac)+1/(c+1+ac)(3)
[(1)+(2)+(3)]/3
=[1+1+1]/3=1