Inequality with numbers 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.

Hint: Note that $2b^2-4ab=2b^2-4ab+2a^2-2a^2=2(a-b)^2-2a^2$. (We completed the square.)  That will tell you what, given $a$, is the optimal choice for $b$.
