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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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    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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    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