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HMMT November

Algebra Difficulty 4.7 AIME Find the answer

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

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

We have ab+ca+b+c=ab−a−ba+b+c+1\frac{a b+c}{a+b+c}=\frac{a b-a-b}{a+b+c}+1 We note that ab−a−ba+b+c<0⇔(a−1)(b−1)<1\frac{a b-a-b}{a+b+c}<0 \Leftrightarrow(a-1)(b-1)<1, which only occurs when either a=1a=1 or b=1b=1. Without loss of generality, let a=1a=1. Then, we have a value of −1b+c+a+1\frac{-1}{b+c+a}+1 We see that this is minimized when bb and cc are also minimized (so b=c=1b=c=1 ), for a value of 23\frac{2}{3}.

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