09/09/06 21:24:18
>>434
直線の傾きを -m とおく。(m>0)
PQ = (b/m)√(1+m^2),
PR = a･√(1+m^2),
(PQ)^2 = (PQ + PR)^2 = (1+m^2)(a + b/m)^2
= a^2 + a(am^2 + b/m + b/m) + b{am + am + b/(m^2)} + b^2
≧ a^2 + 3a^(4/3)^2･b^(2/3) + 3a^(2/3)･b^(4/3) + b^2 (←相加･相乗平均)
= {a^(2/3)^2 + b^(2/3)}^3,
等号成立は m = (b/a)^(1/3) のとき。