Question

在△ABC中,求证:cosA/a+cosB/b+cosC/c=(a²+b²+c²)/2ab

Answer 1

由余弦定理：c^2 = a^2 + b^2 - 2abcosC 变形,得：cosC = ( a^2 + b^2 - c^2 ) / 2ab 同理：cosB = ( a^2 + c^2 - b^2 ) / 2ac cosA = ( b^2 + c^2 - a^2 ) / 2bc 所以：cosA/a + cosB/b + cosC/c = ( b^2 + c^2 - a^2 ) / 2abc + ( a^2 + c^2 - b^2 ) / 2abc + ( a^2 + b^2 - c^2 ) / 2abc = (a^2 + b^2 + c^2) / 2abc >= (ab + bc + ac) / 2abc = 1/2 (1/a + 1/b + 1/c) 取等号的条件是 a = b = c