reserve a,b,c,d,x,j,k,l,m,n for Nat,
  p,q,t,z,u,v for Integer,
  a1,b1,c1,d1 for Complex;

theorem
  n>0 & a>b implies (a|^n+b|^n)*(a-b) <= a|^(n+1) - b|^(n+1)
  proof
    assume
    A1: n>0 & a > b; then
    A2: 0 <= a|^m - b|^m by Lm3a,XREAL_1:48;
    A2b: a|^(n+1)=a*a|^n & b|^(n+1)=b*b|^n & a|^(m+1)=a*a|^m & b|^(m+1)=b*b|^m
    by NEWTON:6;
    consider m such that
    A3: n = 1 + m by A1,NAT_1:14,NAT_1:10;
    A4: (a|^n+b|^n)*(a-b) = a|^(n+1) - b|^(n+1) + a*b|^n - b*a|^n by A2b
    .= a|^(n+1) - b|^(n+1) + a*(b|^(m+1)) - b*(a*a|^m) by NEWTON:6,A3
    .= a|^(n+1) - b|^(n+1) + a*(b*b|^m) - b*(a*a|^m) by NEWTON:6
    .= a|^(n+1) - b|^(n+1) - a*b*(a|^m - b|^m);
    a|^m - b|^m in NAT by A2,INT_1:3;
    hence thesis by A4,XREAL_1:43;
  end;
