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=1 implies (1/a^2-1)*(1/b^2-1)>=9
proof
  assume
A1: a+b=1;
  sqrt(a*b)>0 by SQUARE_1:25;
  then 1^2>=(2*sqrt(a*b))^2 by A1,SIN_COS2:1,SQUARE_1:15;
  then 1>=2*2*(sqrt(a*b))^2;
  then 1>=4*(a*b) by SQUARE_1:def 2;
  then 8/1<=8/(4*a*b) by XREAL_1:118;
  then 8<=8/(4*(a*b));
  then 8<=8/4/(a*b) by XCMPLX_1:78;
  then
A2: 8+1<=2/(a*b)+1 by XREAL_1:7;
  (1/a^2-1)*(1/b^2-1) =(1-1*a^2)/a^2*(1/b^2-1) by XCMPLX_1:126
    .=(1-1*a^2)/a^2*((1-1*b^2)/b^2) by XCMPLX_1:126
    .=((1-a)*(1+a)*((1-b)*(1+b)))/(a^2*b^2) by XCMPLX_1:76;
  then (1/a^2-1)*(1/b^2-1) =((a*b)*((1+a)*(1+b)))/((a*b)*(a*b)) by A1
    .=((a*b)/(a*b))*(((1+a)*(1+b))/(a*b)) by XCMPLX_1:76
    .=1*(((1+a)*(1+b))/(a*b)) by XCMPLX_1:60
    .=(1+(a+b)+a*b)/(a*b)
    .=2/(a*b)+(a*b)/(a*b) by A1,XCMPLX_1:62
    .=2/(a*b)+1 by XCMPLX_1:60;
  hence thesis by A2;
end;
