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 & b+c>a & a+c>b implies 1/(a+b-c)+1/(b+c-a)+1/(c+a-b)>=9/(a+b+c)
proof
  assume that
A1: a+b>c and
A2: b+c>a and
A3: a+c>b;
A4: a+c-b>0 by A3,XREAL_1:50;
  set f=c+a-b;
  set e=b+c-a;
  set d=a+b-c;
A5: b+c-a>0 by A2,XREAL_1:50;
A6: a+b-c>0 by A1,XREAL_1:50;
  then
A7: d+e+f>=3*(3-root(d*e*f)) by A5,A4,SERIES_3:15;
  1/d+1/e+1/f>=3*(3-root((1/d)*(1/e)*(1/f))) by A6,A5,A4,SERIES_3:15;
  then
A8: 1/d+1/e+1/f>=3*(3-root((1/(d*e*f)))) by Lm3;
A9: 3-root(d*e*f)>=0 by A6,A5,A4,POWER:7;
  3-root(1/(d*e*f))>=0 by A6,A5,A4,POWER:7;
  then
  (d+e+f)*(1/d+1/e+1/f)>=(3*(3-root(d*e*f)))*(3*(3-root((1/(d*e*f))))) by A7,A9
,A8,XREAL_1:66;
  then (d+e+f)*(1/d+1/e+1/f)>=9*((3-root(d*e*f))*(3-root((1/(d*e*f)))));
  then (d+e+f)*(1/d+1/e+1/f)>=9*(3-root((d*e*f)*(1/(d*e*f)))) by A6,A5,A4,
POWER:11;
  then (d+e+f)*(1/d+1/e+1/f)>=9*(3-root((d*e*f)/((d*e*f)/1))) by XCMPLX_1:79;
  then (d+e+f)*(1/d+1/e+1/f)>=9*(3-root(1)) by A6,A5,A4,XCMPLX_1:60;
  then (d+e+f)*(1/d+1/e+1/f)>=9*1 by POWER:6;
  hence thesis by XREAL_1:79;
end;
