>>144

[a+b] = [a] + [b] + [{a}+{b}] ≧ [a] + [b],
j≧2 のとき
S_j = (j-1)・[jx] - 2Σ(k=1,j-1) [kx]
  = Σ(k=1,j-1) ( [jx] - [kx] - [(j-k)x] )
  ≧ 0,

f(x) = [nx] - Σ(k=1,n) [kx]/k
 = Σ(j=2,n-1) (1/j - 1/(j+1))・S_j + (1/n)・S_n
 ≧ 0,