- 833 名前：１３２人目の素数さん mailto:sage [2009/03/17(火) 00:00:13 ]
- >>830
(1)
H = 3/(1/x + 1/y + 1/z), G = (xyz)^(1/3), A = (x+y+z)/3, Q = √{(x^2 + y^2 + z^2)/3},
Q^2 - A^2 = (x^2 +y^2 +z^2)/3 - (1/9)(x+y+z)^2 = (1/9){(x-y)^2 + (y-z)^2 + (z-x)^2} ≧ 0,
A^3 - G^3 = {(x+y+z)/3}^3 -xyz = {(x+y+7z)/2}(x-y)^2 + {(7x+y+z)/2}(y-z)^2 + {(x+7y+z)/2}(z-x)^2 ≧ 0,
(1/H)^3 - (1/G)^3 = {[(1/x)+(1/y)+(1/z)]/3}^3 - 1/(xyz) = {(x'+y'+z')/3}^3 - x'y'z' ≧ 0,
∴ H ≦ G ≦ A ≦ Q,
(2) y=z=1 の場合を考えると
H = 3x/(1+2x), G = x^(1/3), A = (2+x)/3, Q = √{(2+x^2)/3},
x<1 のとき G-H > A-G > Q-A,
x>1 のとき G-H < A-G < Q-A,
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