HMMT-Nov 2011 guts, Problem 8

Let $a, b, c$ be not necessarily distinct integers between 1 and 2011, inclusive. Find the smallest possible value of $\frac{a b+c}{a+b+c}$.

Solution: We have \[ \frac{a b+c}{a+b+c}=\frac{a b-a-b}{a+b+c}+1 \] We note that $\frac{a b-a-b}{a+b+c}<0 \Leftrightarrow(a-1)(b-1)<1$, which only occurs when either $a=1$ or $b=1$. Without loss of generality, let $a=1$. Then, we have a value of \[ \frac{-1}{b+c+a}+1 \] We see that this is minimized when $b$ and $c$ are also minimized (so $b=c=1$ ), for a value of $\frac{2}{3}$. $\fbox{\frac{2}{3}}$.

Answer: \frac{2}{3}