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 implies a|^(n+1)+b|^(n+1)>=a|^n*b+a*b|^n
proof
  assume
A1: n>=1;
  per cases;
  suppose
A2: a-b>0;
    then a-b+b>0+b by XREAL_1:8;
    then a|^n>b|^n by A1,PREPOWER:10;
    then a|^n-b|^n>0 by XREAL_1:50;
    then (a-b)*(a|^n-b|^n)>0 by A2;
    then a*(a|^n)-a*(b|^n)-b*(a|^n)+b*(b|^n)>0;
    then a|^(n+1)-a*(b|^n)-b*(a|^n)+b*(b|^n)>0 by NEWTON:6;
    then a|^(n+1)-a*(b|^n)-b*(a|^n)+b|^(n+1)>0 by NEWTON:6;
    then
    (a|^(n+1)+b|^(n+1)-a*(b|^n)-b*(a|^n))+(a*(b|^n)+b*(a|^n))> 0+(a*(b|^n
    )+b*(a|^n)) by XREAL_1:8;
    hence thesis;
  end;
  suppose
A3: a-b=0;
    then a|^n*b+a*b|^n =a|^(n+1)+(a|^n*a) by NEWTON:6
      .=a|^(n+1)+a|^(n+1) by NEWTON:6
      .=2*(a|^(n+1));
    hence thesis by A3;
  end;
  suppose
A4: a-b<0;
    then (a-b)+b<0+b by XREAL_1:8;
    then a|^n<b|^n by A1,PREPOWER:10;
    then a|^n-b|^n<0 by XREAL_1:49;
    then (a-b)*(a|^n-b|^n)>0 by A4;
    then a*(a|^n)-a*(b|^n)-b*(a|^n)+b*(b|^n)>0;
    then a|^(n+1)-a*(b|^n)-b*(a|^n)+b*(b|^n)>0 by NEWTON:6;
    then a|^(n+1)-a*(b|^n)-b*(a|^n)+b|^(n+1)>0 by NEWTON:6;
    then
    (a|^(n+1)+b|^(n+1)-a*(b|^n)-b*(a|^n))+(a*(b|^n)+b*(a|^n))> 0+(a*(b|^n
    )+b*(a|^n)) by XREAL_1:8;
    hence thesis;
  end;
end;
