由于正数A,B,C成等差数列,则 2B=A+C
若1/(√B+√C),1/(√C+√A),1/(√A+√B)成等差数列
则有 2/(√C+√A)=1/(√B+√C)+1/(√A+√B)
=(2√B+√A+√C)/(B+√AB+√BC+√AC)
则有2(B+√AB+√BC+√AC)=(√C+√A)(2√B+√A+√C)
=2(B+√AB+√BC+√AC)
由于以上各步均可逆,故命题得证
由于正数A,B,C成等差数列,则 2B=A+C
若1/(√B+√C),1/(√C+√A),1/(√A+√B)成等差数列
则有 2/(√C+√A)=1/(√B+√C)+1/(√A+√B)
=(2√B+√A+√C)/(B+√AB+√BC+√AC)
则有2(B+√AB+√BC+√AC)=(√C+√A)(2√B+√A+√C)
=2(B+√AB+√BC+√AC)
由于以上各步均可逆,故命题得证