theorem
  for a,b be non zero Nat st c >= a+b holds
    c|^(k+1)*(a+b) > a|^(k+2) + b|^(k+2)
  proof
    let a,b be non zero Nat;
    A1: a|^(1) + b|^(1) <= (a+b)|^(1) implies
    a|^((k+1)+1) + b|^((k+1)+1) < (a+b)|^((k+1)+1) by NEWTON01:18;
    assume c >= a+b; then
    c|^(k+1) >= (a+b)|^(k+1) by NEWTON02:40; then
    c|^(k+1)*(a+b) >= (a+b)|^(k+1)*(a+b) by XREAL_1:64; then
    c|^(k+1)*(a+b) >= (a+b)|^((k+1)+1) by NEWTON:6;
    hence thesis by A1,XXREAL_0:2;
  end;
