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
  a+b+c = 1 implies 1/a+1/b+1/c>=9
proof
  assume
A1: a+b+c = 1;
A2: 3-root(a*b*c)>=0 by POWER:7;
  1/a+1/b+1/c>=3*(3-root((1/a)*(1/b)*(1/c))) by SERIES_3:15;
  then
A3: 1/a+1/b+1/c>=3*(3-root((1/(a*b*c)))) by Lm3;
A4: 3-root((1/(a*b*c)))>=0 by POWER:7;
  a+b+c>=3*(3-root(a*b*c)) by SERIES_3:15;
  then
  (a+b+c)*(1/a+1/b+1/c)>=(3*(3-root(a*b*c)))*(3*(3-root((1/(a*b*c))))) by A2,A3
,A4,XREAL_1:66;
  then (1/a+1/b+1/c)>=9*((3-root(a*b*c))*(3-root((1/(a*b*c))))) by A1;
  then (1/a+1/b+1/c)>=9*(3-root((a*b*c)*(1/(a*b*c)))) by POWER:11;
  then (1/a+1/b+1/c)>=9*(3-root((a*b*c)/((a*b*c)/1))) by XCMPLX_1:79;
  then (1/a+1/b+1/c)>=9*(3-root(1)) by XCMPLX_1:60;
  then (1/a+1/b+1/c)>=9*1 by POWER:6;
  hence thesis;
end;
