2018 USAMO Problems/Problem 2

Revision as of 19:45, 19 March 2022 by Wzs26843545602 (talk | contribs) (Solution)
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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 $xyz =1.$


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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