It seems its a simple question, but I am confused.
Let a be natural number and let b be some number $1\le b\le a$. Find an upper bound for $ \frac{a^2+2b^2-4ab-a}{a(a-1)}. $
I've got $ \frac{a^2+2b^2-4ab-a}{a(a-1)}= \frac{a+2b^2/a-4b-1}{(a-1)}\leq 3 $ But I am not sure if its true.
Thank you.