Let the nos. be #a and b; a,b >0#.
Then, in the Usual Notations,
#A=(a+b)/2, G=sqrt(ab), and, H=(2ab)/(a+b)...............(star)#.
Given Data#rArr A:G=5:4............(1), and, |G-H|=4/5...........(2)#
Knowing that, #A>=G>=H", we rewrite (2) as, "G-H=4/5.....(2')#.
#(1) rArr (a+b)/(2sqrt(ab))=5/4 rArr 2(a+b)=5sqrt(ab)#
#rArr 4(a^2+2ab+b^2)-25ab=0", i.e., "4a^2-17ab+4b^2=0#
#rArr(a-4b)(4a-b)=0 rArr a=4b, or, b=4a#
Case : 1 : a=4b
Then, #G=sqrt(ab)=2b, and, H=(2*4b*b)/(4b+b)=8/5b#
Hence, by #(2'), 2b-8/5b=4/5 rArr 2/5b=4/5 rArr b=2#
Thus, in this Case, the Nos. are, #8,and, 2#.
Case : 2 : b=4a
We simply notice that the roles of #a and b# have interchanged, so,
we immediately jump to the conclusion that, in this Case, the reqd.
Nos. would be #2, and, 8#.
Enjoy Maths.!
