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Maximize the value of the function $$ z=\frac{ab+c}{a+b+c}, $$ where $a,b,c$ are natural numbers and are all lesser than 2010 and not necessarily distinct from each other. Please provide a proof, and if possible a general technique. Thank you.

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    Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level.2012-12-16
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    Thank you sir. I will keep that in mind in the future.2012-12-17

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$$z=\frac{ab+c}{a+b+c}=\frac{ab-a-b}{a+b+c}+1$$

Clearly, this will be maximum if $c$ minimum $=1$

So, $$z\le \frac{ab-a-b}{a+b+1}+1=\frac{ab+1}{a+b+1} $$

$\frac{a_1b_1+1}{a_1+b_1+1}$ will be greater than $\frac{a_2b_2+1}{a_2+b_2+1}$

if $(a_1a_2-1)(b_1-b_2)+(b_1b_2-1)(a_1-a_2)+a_1b_1-a_2b_2>0$

if $(a_1a_2-1)(b_1-b_2)+(b_1b_2-1)(a_1-a_2)+(a_1-a_2)b_1+a_2(b_1-b_2)>0$

if $(a_1a_2-1+a_2)(b_1-b_2)+(b_1b_2-1+b_1)(a_1-a_2)>0$

Clearly, $\frac{ab+1}{a+b+1} $ increases with the increment of $a,b$ or both.

So, $\frac{ab+1}{a+b+1} $ will be maximum if $a,b$ are maximum.

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    Thank you. So in a way we could use multivariable calculus to solve this question.2012-12-17
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    @Noel, if you are talking about using partial derivative, I'm not sure how to apply the given constraint on the partial derivatives.2012-12-17
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    Well I think we can just prove it to be increasing with the values of a,b and decreasing with c. That is what you have done.2012-12-22