>>21
[前スレ.360(2)]
　Σ[k=1,m] x_k↑ = s↑, (m-2)Σ[k=1,m] |x_k| +|s| =L,　Σ[1≦i<j≦m] |x_i +x_j| =R
とおく。まず補題として
　D = (m-2)Σ[k=1,m] |x_k|^2 +|s|^2 - Σ[1≦i<j≦m] |x_i +x_j|^2 =0.
を示すのだが、これは内積を使えば簡単である。

次に L-R ≧0 を示そう。
(L-R){ Σ[k=1,m] |x_k| +|s| } = (m-2){Σ[k=1,m] |x_k|}^2 +(m-1){Σ[k=1,m] |x_k|}|s| + |s|^2 -R{……}
　= 2(m-2)納i<j] |x_i||x_j| +(m-1){Σ[k=1,m] |x_k|}|s| - R{……} + (m-2)Σ[k=1,m] |x_k|^2 + |s|^2
　= 2(m-2)納i<j] |x_i||x_j| +(m-1){Σ[k=1,m] |x_k|}|s| - R{……} + Σ[i<j] |x_i +x_j|^2 +D
　= 納i<j] {(|x_i|+|x_j|)(納k≠i,j] |x_k| +|s|) - {……}|x_i+x_j| + |x_i+x_j|^2 } +D
　= 納i<j] {|x_i|+|x_j|-|x_i +x_j|}･{納k≠i,j] |x_k| +|s| -|x_i+x_j|} +D
　≧ D =0.
{……} >0 だから L-R ≧0.

*)　納k≠i,j] |x_k| +|s| -|x_i+x_j|
　　= (納k≠i,j] |x_k| -|納k≠i,j] x_k| ) + (|s -x_i -x_j| +|s| -|x_i +x_j|) ≧0

参考書[3] の p.34