09/06/13 00:21:06
>>957 上
a+b+c=s, ab+bc+ca=t, abc=u とおくと、
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-c)/{(b-c)(b-a)} = -1/(a-b) - 1/(b-c) より
(左辺)*(a-c) = {1/(a-b)}(1/√a - 1/√b) + {1/(b-c)}(1/√c - 1/√b)
= - 1/(a√b + b√a) + 1/(c√b + b√c) > 0, (← a>c)
>>958
対数を考えれ。チェビシェフより
Σ[k=1,n] (a_k)log(a_k) ≧ {Σ[i=1,n] log(a_i)}(Σ[j=1,n] a_j)/n,