- 107 名前：１３２人目の素数さん mailto:sage [2007/07/14(土) 05:35:41 ]
- >102
(xy^2)^(1/3) =Z, (yz^2)^(1/3) =X, (zx^2)^(1/3) =Y とおくと
(X+Y+Z)/3 ≧ XYZ = xyz,
g(t) = (t^n +1)^(1/n) とおくと
g'(t) = t^(n-1) /（t^n +1)^(1 -1/n) >0, (単調増加)
g"(t) = (n-1)t^(n-2) / (t^n +1)^(2 -1/n) >0, (下に凸)
(左辺)^3 = {g(X) + g(Y) + g(Z)}/3 ≧ g((X+Y+Z)/3) ≧ g((XYZ)^(1/3)) = g((xyz)^(1/3)) = (右辺)^3,
