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| | ==Solution== | | ==Solution== |
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| − | The only such function is <math>f(x)=\frac13</math>.
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| − | Letting <math>x=y=z=1</math> gives <math>3f(2)=1</math>, hence <math>f(2)=\frac13</math>. Now observe that even if we fix <math>x+\frac1y=y+\frac1z=2</math>, <math>z+\frac1x</math> is not fixed. Specifically,
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| − | <cmath>y=\frac1{2-x}</cmath>
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| − | <cmath>z=\frac1{2-y}=\frac{2-x}{3-2x}</cmath>
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| − | <cmath>z+\frac1x=\frac12+\frac1x+\frac1{2(3-2x)}</cmath>
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| − | This is continuous on the interval <math>x\in\left(0,\frac32\right)</math> and has an asymptote at <math>x=\frac32</math>. Since it takes the value 2 when <math>x=1</math>, it can take on all values greater than or equal to 2. So for any <math>a\ge2</math>, we can find <math>x</math> such that <math>2f(2)+f(x)=1</math>. Therefore, <math>f(x)=\frac13</math> for all <math>x\ge2</math>.
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| − | Now, for any <math>0<k<2</math>, if we let <math>x=\frac k2</math>, <math>y=\frac1x</math>, and <math>z=1</math>, then <math>f(k)+2f\left(1+\frac2k\right)=1</math>. Since <math>1+\frac2k>2</math>, <math>f\left(1+\frac2k\right)=\frac13</math>, hence <math>f(k)=\frac13</math>. Therefore, <math>f(x)=\frac13</math> for all <math>x\ge0</math>.
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| − | -- wzs26843645602
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| − | ==Solution 2==
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| − | Make the substitution <math>x = \frac{a}{b},</math> <math>y = \frac{b}{c},</math> <math>z = \frac{c}{a},</math> so that the given equation becomes <cmath>f\bigg(\frac{a+b}{c} \bigg) + f \bigg(\frac{a+c}{b} \bigg) + f\bigg(\frac{b+c}{a} \bigg) = 1</cmath> for all <math>a, b, c \in \mathbb{R}_{>0}.</math> Now, let <cmath>\bigg(X, Y, Z \bigg) = \bigg(\frac{a+b}{c}, \ \frac{a+c}{b}, \ \frac{b+c}{a} \bigg).</cmath> We will show that if we fix <math>Y</math> and <math>Z,</math> <math>X</math> can still vary to be any positive real number. Notice that <math>(a,b,c) = \bigg((X+1)(Y+1), \ (X+1)(Z+1), \ (Y+1)(Z+1) \bigg)</math> will be a solution to the system of equations <cmath>\begin{cases} \frac{a+b}{c} = X \\ \frac{a+c}{b} = Y \\ \frac{b+c}{a} = Z, \\ \end{cases},</cmath> and so even if we fix <math>Y</math> and <math>Z</math> this system of equations will have a solution in <math>(a,b,c)</math> for any real <math>X \in (0, \infty).</math> Now, for any real <math>X \in (0, \infty),</math> and for a fixed <math>Y</math> and <math>Z,</math> we have <math>f(X) + f(Y) + f(Z)</math> equals <math>1</math>, which implies that <math>f</math> is a constant function over its domain <math>(0, \infty).</math> However, letting <math>f(x) = k</math> for all real <math>x > 0,</math> we see that <cmath>f(X) + f(Y) + f(Z) = 1, \ \ \implies \ \ 3k = 1,</cmath> so the only possible value for <math>k</math> is <math>\frac{1}{3}.</math> Thus, the only possible function is <math>\boxed{f(x) = \frac{1}{3}},</math> which obviously satisfies all necessary conditions.
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| − | ~ Professor-Mom
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such that
![\[f\left(x+\frac{1}{y}\right)+f\left(y+\frac{1}{z}\right) + f\left(z+\frac{1}{x}\right) = 1\]](http://latex.artofproblemsolving.com/8/9/4/8945fb40f421318fdb191894bc65db33f0253cd0.png)
for all 
with 
