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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.

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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 $b$.

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    @HagenvonEitzen: I initially had misread the question as asking for the *minimum*, caught it on editing.2012-09-18