2018 USAMO Problems/Problem 2

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 $x\ge0$ .

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2018 USAMO Problems/Problem 2

Problem 2

Solution