由余弦定理,a² = b²+c²-2bc·cosA,b² = c²+a²-2ca·cosB,c² = a²+b²-2ab·cosC.
相加得a²+b²+c² = 2(ab·cosC+bc·cosA+ca·cosB) = 2abc(cosA/a+cosB/b+cosC/c).
由正弦定理,a = 2R·sinA,b = 2R·sinB,c = 2R·sinC.
代入得a²+b²+c² = abc/R·(cosA/sinA+cosB/sinB+cosC/sinC) = abc/R·(cotA+cotB+cotC).
于是1/2·(a²+b²+c²)(1/(ab)+1/(bc)+1/(ca)) = (a²+b²+c²)(a+b+c)/(2abc)
= (a+b+c)/(2R)·(cotA+cotB+cotC) = (a/(2R)+b/(2R)+c/(2R))(cotA+cotB+cotC)
= (sinA+sinB+sinC)(cotA+cotB+cotC).