Question

已知 abc=1,求证：1/ab+a+1 + 1/bc+b+1 + 1/ca+c+1=1

Answer 1

abc=1

所以

b=1/ac

ab=1/c

bc=1/a

所以左边

=1/(1/c+a+1)+1/(1/a+1/ac+1)+1/(ac+c+1)

第一个式子上下乘c

第二个式子上下乘ac

=c/(ac+c+1)+ac/(ac+c+1)+1/(ac+c+1)

=(ac+c+1)/(ac+c+1)

=1=右边