已知:abc=1,试说明1/(ab+a+1)+1/(bc+b+1)+1/(ac+c+1)=1

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已知:abc=1,试说明1/(ab+a+1)+1/(bc+b+1)+1/(ac+c+1)=1
已知:abc=1,试说明1/(ab+a+1)+1/(bc+b+1)+1/(ac+c+1)=1

已知:abc=1,试说明1/(ab+a+1)+1/(bc+b+1)+1/(ac+c+1)=1
证明:[1/(ab+a+1)]+[1/(bc+b+1)]+[1/(ac+c+1)] =[1/(ab+a+abc)]+[1/(bc+b+1)]+[1/(ac+c+abc)] =[1/a(b+1+bc)]+[1/(b+1+bc)]+[1/c(a+1+ab)] =[1/a(b+1+bc)]+[1/(b+1+bc)]+[1/ac(b+1+bc)] =1/(b+1+bc)*[(1/a)+1+(1/ac)] =1/(b+1+bc)*[bc+1+b] =1