04/09/16 11:36:25
>392
【類題】n≧3, a_1~a_nのk次の対称式をS_k とおくと、
(S_1)^3 - Σ[k=1~n](a_k)^3 ≧ {6(n+1)/(n-2)}S_3.
(略証)
左辺 - 右辺 = 3(S_1・S_2 - S_3) - {6(n+1)/(n-2)}S_3
= [3/(n-2)]{(n-2)S_1・S_2 - 3nS_3}
(n-2)S_2・S_1 -3nS_3
= (n-2)Σ[i<j] a_i・a_j Σ[k=1~n] a_k - 3nΣ[i<j<k] a_i・a_j・a_k
= (n-2)Σ[i<j] a_i・a_j・(a_i+a_j) - 6Σ[i<j<k] a_i・a_j・a_k
= Σ[i<j] {Σ[k≠i,j]a_k}{(a_i)^2 +(a_j)^2} - Σ[i<j] {Σ[k≠i,j]a_k}2a_i・a_j
= Σ[i<j] {Σ[k≠i,j]a_k}(a_i-a_j)^2 ≧ 0.
ぬるぽ