(In reply to question re: Possible solution
Answer: I was called away half-way through before I could finish.
c = (ab/(2a-b))
Now we have a^2+b^2= (ab/(2a-b))^2+1
a cannot be 1, for then b=c.
Say a is much larger than b:
Then (a/(2a))^2 is around (a^2)-1 involving the contradiction that a is less than 2.
Say b is much larger than a:
Then (b/(-b))^2 is around (b^2)-1 involving the contradiction that b is less than 2.
Say a and b are around the same size
(a^2/(a))^2 is around (2a^2-1)
and a is still less than 2.
So a can't be 1, yet can't be large enough to be greater than 1.
Posted by broll
on 2015-06-23 00:45:51