17/07/13 18:37:29.12 oVTfqBd/.net
>>72
A.422
Σ[i=1,n] x(i) = x(n+1) = S とおく。
Σ[i=1,n] x(i)^2 ≧ SS/n,
y=√x は上に凸だから
(左辺)^2 ≦ n｛ Σ[i=1,n] x(i) [S -x(i)] }
= n｛ SS -Σ[i=1,n] x(i)^2 }
≦ n (SS - SS/n)}
= (n-1) SS,
(右辺)^2 = SΣ[i=1,n] [S - x(i)]
= S (n S - S)
= (n-1) SS,