Baltic Way 2024 · Problem 2

Note that $\frac{1}{1+a+c a}=b c \cdot \frac{1}{1+c+b c}$ since $a b c=1$. Similarly,

$$\frac{1}{1+b+a b}=a c \cdot \frac{1}{1+a+c a}=c \cdot \frac{1}{1+c+b c}$$

So the initial equality becomes $\frac{b c f(a)+c f(b)+f(c)}{1+c+b c}=1$ which yields

$$b c f\left(\frac{1}{b c}\right)+c f(b)+f(c)=1+c+b c$$

Taking $a=b=c=1$ in (8) gives $f(1)+f(1)+f(1)=3$ which implies $f(1)=1$. Using this fact after substituting $c=1$ into 8 yields $b f\left(\frac{1}{b}\right)+f(b)=1+b$, so $b f\left(\frac{1}{b}\right)=1+b-f(b)$ for all $b \in \mathbb{R}^{+}$. Applying this in (8) gives $1+b c-f(b c)+c f(b)+f(c)=1+c+b c$, so

$$c f(b)+f(c)=c+f(b c)$$

Swapping $b$ and $c$ here gives

$$b f(c)+f(b)=b+f(b c)$$

Subtracting the last equality from the second last one and rearranging the terms gives

$$c f(b)-f(b)+f(c)-b f(c)=c-b$$

Substituting $c=2$ into (9) gives $f(b)+f(2)-b f(2)=2-b$, so $f(b)=b(f(2)-1)+2-f(2)$. Denoting $f(2)-1=k$, we get $f(b)=k b+1-k$ for all $b \in \mathbb{R}^{+}$. Note that if $k<0$, then for large enough $b$ the value of $f(b)$ would become negative. If $k>1$, then for small enough $b$ the value of $f(b)$ would become negative. Therefore $k \in[0,1]$. After substituting $f(x)=k x+1-k$ into 8e can see that it is satisfied. Hence this function satisfies the original equality for all $a, b, c \in \mathbb{R}^{+}$such that $a b c=1$.