09/04/18 10:54:55
>>863
⊿ = (1/2)ab･sin(C) = (1/2)bc･sin(A) = (1/2)ca･sin(B),
でも解けるお。
⊿ = (1/2)(abc)^(2/3)･{sin(A)sin(B)sin(C)}^(1/3)
≦ (1/2)(abc)^(2/3)･{sin(A)+sin(B)+sin(C)}/3 (相加･相乗平均)
≦ (1/2)(abc)^(2/3)･sin((A+B+C)/3) (0～πで上に凸)
= (1/2)(abc)^(2/3)･sin(π/3)
= (√3)/4･(abc)^(2/3), (等号成立はA=B=C(正三角形)のとき.)
⊿ '≦ (√3)/4･(1/abc)^(2/3),
辺々掛ける。