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 $x\ge0$ .

-- wzs26843645602

Solution 2

Make the substitution $x = \frac{a}{b},$ $y = \frac{b}{c},$ $z = \frac{c}{a},$ so that the given equation becomes $f\bigg(\frac{a+b}{c} \bigg) + f \bigg(\frac{a+c}{b} \bigg) + f\bigg(\frac{b+c}{a} \bigg) = 1$ for all $a, b, c \in \mathbb{R}_{>0}.$ Now, let $\bigg(X, Y, Z \bigg) = \bigg(\frac{a+b}{c}, \ \frac{a+c}{b}, \ \frac{b+c}{a} \bigg).$ We will show that if we fix $Y$ and $Z,$ $X$ can still vary to be any positive real number. Notice that $(a,b,c) = \bigg((X+1)(Y+1), \ (X+1)(Z+1), \ (Y+1)(Z+1) \bigg)$ will be a solution to the system of equations $\begin{cases} \frac{a+b}{c} = X \\ \frac{a+c}{b} = Y \\ \frac{b+c}{a} = Z, \\ \end{cases},$ and so even if we fix $Y$ and $Z$ this system of equations will have a solution in $(a,b,c)$ for any real $X \in (0, \infty).$ Now, for any real $X \in (0, \infty),$ and for a fixed $Y$ and $Z,$ we have $f(X) + f(Y) + f(Z)$ equals $1$ , which implies that $f$ is a constant function over its domain $(0, \infty).$ However, letting $f(x) = k$ for all real $x > 0,$ we see that $f(X) + f(Y) + f(Z) = 1, \ \ \implies \ \ 3k = 1,$ so the only possible value for $k$ is $\frac{1}{3}.$ Thus, the only possible function is $\boxed{f(x) = \frac{1}{3}},$ which obviously satisfies all necessary conditions.

~ Professor-Mom

Page

Toolbox

Search

2018 USAMO Problems/Problem 2

Problem 2

Solution

Solution 2