reserve a,b,c,d for positive Real,
  m,u,w,x,y,z for Real,
  n,k for Nat,
  s,s1 for Real_Sequence;

theorem
  n>=1 & k>=1 implies (a|^k+b|^k)*(a|^n+b|^n)<=2*(a|^(k+n)+b|^(k+n))
proof
  assume that
A1: n>=1 and
A2: k>=1;
A3: (a|^k+b|^k)*(a|^n+b|^n)-2*(a|^(k+n)+b|^(k+n)) =a|^k*a|^n+a|^k*b|^n+b|^k*
  a|^n+b|^k*b|^n-2*a|^(k+n)-2*(b|^(k+n))
    .=a|^(k+n)+a|^k*b|^n+b|^k*a|^n+b|^k*b|^n-2*a|^(k+n)-2*(b|^(k+n)) by
NEWTON:8
    .=a|^(k+n)+a|^k*b|^n+b|^k*a|^n+b|^(k+n)-2*a|^(k+n)-2*(b|^(k+n)) by NEWTON:8
    .=a|^k*b|^n+b|^k*a|^n-a|^(k+n)-(b|^(k+n))
    .=a|^k*b|^n+b|^k*a|^n-a|^k*a|^n-(b|^(k+n)) by NEWTON:8
    .=a|^k*b|^n+b|^k*a|^n-a|^k*a|^n-b|^k*b|^n by NEWTON:8
    .=(a|^n-b|^n)*(b|^k-a|^k);
  per cases;
  suppose
    a-b>0;
    then
A4: a-b+b>0+b by XREAL_1:8;
    then a|^k>b|^k by A2,PREPOWER:10;
    then
A5: b|^k-a|^k<0 by XREAL_1:49;
    a|^n>b|^n by A1,A4,PREPOWER:10;
    then a|^n-b|^n>0 by XREAL_1:50;
    hence thesis by A3,A5,XREAL_1:48;
  end;
  suppose
    a-b=0;
    then a|^n=b|^n;
    hence thesis by A3;
  end;
  suppose
    a-b<0;
    then
A6: a-b+b<0+b by XREAL_1:8;
    then a|^k<b|^k by A2,PREPOWER:10;
    then
A7: b|^k-a|^k>0 by XREAL_1:50;
    a|^n<b|^n by A1,A6,PREPOWER:10;
    then a|^n-b|^n<0 by XREAL_1:49;
    hence thesis by A3,A7,XREAL_1:48;
  end;
end;
