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Difference between revisions of "2018 USAMO Problems/Problem 2"

(Created page with "==Problem 2== Find all functions <math>f:(0,\infty) \rightarrow (0,\infty)</math> such that <cmath>f\left(x+\frac{1}{y}\right)+f\left(y+\frac{1}{z}\right) + f\left(z+\frac{1}...")
 
(→‎Solution)
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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

Revision as of 18:45, 19 March 2022

Problem 2

Find all functions $f:(0,\infty) \rightarrow (0,\infty)$ such that

\[f\left(x+\frac{1}{y}\right)+f\left(y+\frac{1}{z}\right) + f\left(z+\frac{1}{x}\right) = 1\] for all $x,y,z >0$ with $xyz =1.$


Solution

The only such function is $f(x)=\frac13$.

Letting $x=y=z=1$ gives $3f(2)=1$, hence $f(2)=\frac13$. Now observe that even if we fix $x+\frac1y=y+\frac1z=2$, $z+\frac1x$ is not fixed. Specifically, \[y=\frac1{2-x}\] \[z=\frac1{2-y}=\frac{2-x}{3-2x}\] \[z+\frac1x=\frac12+\frac1x+\frac1{2(3-2x)}\] This is continuous on the interval $x\in\left(0,\frac32\right)$ and has an asymptote at $x=\frac32$. Since it takes the value 2 when $x=1$, it can take on all values greater than or equal to 2. So for any $a\ge2$, we can find $x$ such that $2f(2)+f(x)=1$. Therefore, $f(x)=\frac13$ for all $x\ge2$.

Now, for any $0<k<2$, if we let $x=\frac k2$, $y=\frac1x$, and $z=1$, then $f(k)+2f\left(1+\frac2k\right)=1$. Since $1+\frac2k>2$, $f\left(1+\frac2k\right)=\frac13$, hence $f(k)=\frac13$. Therefore, $f(x)=\frac13$ for all $x\ge0$.

-- wzs26843645602

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