09/06/13 00:42:37
>>957 上
a^3 + b^3 +c^3 -3abc = (a+b+c){(a-b)^2 +(b-c)^2 +(c-a)^2}/2
≧ 3{(a-b)^2 +(b-c)^2 +(c-a)^2}/2 (← a,b,c≧1)
≧ (1/a + 1/b + 1/c){[(a-b)/ab]^2 + [(b-c)/bc]^2 + [(c-a)/ca]^2}/2 (← 1≧1/a,1/b,1/c)
= 1/(a^3) + 1/(b^3) + 1/(c^3) - 3/(abc),
>>957 下
√a = A, √b = B, √c = C とおくと、
(左辺)*(a-c) = (A-C)(A+B+C)/{(A+B)(B+C)ABC},
(左辺) = (A+B+C)/{(A+B)(B+C)(C+A)ABC} >0,