theorem
  n>0 & a<>b implies 2*(a|^n)*(b|^n) < a|^(2*n) + b|^(2*n)
  proof
    A0: (a|^n)|^2 = a|^(2*n) & (b|^n)|^2 = b|^(2*n) by NEWTON:9;
    assume n>0 & a<>b; then
    n>=1 & (a>b or b>a) by NAT_1:14,XXREAL_0:1; then
    a|^n <> b|^n by PREPOWER:10;
    hence thesis by A0,Th55;
  end;
