a1>b1>0; an+1=(an+bn)/2; bn+1=2anbn/(an+bn).
an=2an+1-bn, so an-an+1=an+1-bn; bn+1=anbn/an+1, by substituting 1/an+1 for 2/(an+bn), therefore bn+1/bn=an/an+1 or an+1bn+1=anbn.
When n=1, a2=(a1+b1)/2; b2=b1(a1/a2).
Since a1>b1, let a1=b1+s1 where s1>0, then b1=a1-s1 and a2=(2a1-s1)/2=a1-s1/2, therefore a2<a1 and a1/a2>1, so b2>b1.
an+1/bn+1=[(an+bn)/2]/[2anbn/(an+bn)]=[(an+bn)/2][(an+bn)/(2anbn)]=(an+bn)2/(4anbn).
When n=1, a2/b2=(2a1-s1)2/(4a1(a1-s1))=(4a12-4a1s1+s12)/(4a12-4a1s1)=1+s12/(4a12-4a1s1).
s12/(4a12-4a1s1)>0 so a2/b2>1, a2>b2.
So we have (for n=1), a1>a2>b2>b1, which is the special case of an>an+1>bn+1>bn.
But we need to show that this is the case for general n.
More to follow in due course...