Find the smallest possible value of a fraction

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SUMMARY

The smallest possible value of the fraction $\dfrac{xy+z}{x+y+z}$, where $x, y, z$ are integers between 1 and 2011, is determined to be $\frac{1}{3}$. This conclusion is reached by analyzing the expression and substituting values for $x$, $y$, and $z$. Specifically, setting $x = 1$, $y = 1$, and $z = 1$ yields the minimum value, confirming that the fraction can achieve this lower bound under the given constraints.

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Let $x,\,y,\,z$ be not necessarily distinct integers between 1 and 2011, inclusive. Find the smallest possible value of $\dfrac{xy+z}{x+y+z}$.
 
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anemone said:
Let $x,\,y,\,z$ be not necessarily distinct integers between 1 and 2011, inclusive. Find the smallest possible value of $\dfrac{xy+z}{x+y+z}$.
the smallest possible value of $\dfrac{xy+z}{x+y+z}$.will exist when xy<x+y
if x=1 then y=1,2,3,-----2011
if y=1 then x=1,2,3,-----2011
for $\dfrac {n}{n+1}<\dfrac {n+1}{n+2}$
for all $n\in N$
$\therefore$ the smallest value of $\dfrac{xy+z}{x+y+z}$=$\dfrac{2}{3}$
here $x=y=z=1$
 
$\frac{xy+z}{x+y+z}$
= $1+ \frac{xy-x - y}{x+y+z}$
= $1+ \frac{(x-1)(y-1) -1}{x+y+z}$

(x-1)(y-1) - 1 is positive for all x and y except for x=1 or y=1 ( in the condition x, y <= 2011)

so x =1 , and y = 1

so we get given expression
= $1- \frac{1}{2+z}$
z = 1 shall make it lowest

so x = 1 = y = z shall give the value $\frac{2}{3}$
 
Last edited:

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