Problem 1995

$f(x) = \frac{-1+\sqrt{5}}{2}$ for all $x \in \mathbb{R}_+$, or $f(x) = \frac{x}{x+1}$ for all $x \in \mathbb{R}_+$.

Denote the functional equality by $P(x, y)$. Then $P((w+1)^{-1}, w)$ implies
$$f(w+1) (f(w^{-1}) + 1) = 1. \qquad (1)$$
By $P(xy + x + y, z)$ and (1), we have
$$1 - f((x+1)(y+1)(z+1) - 1) = f\left(\frac{1}{xy + x + y}\right) f\left(\frac{1}{z}\right).$$
Since LHS is symmetric with respect to $x, y, z$, by symmetry, we have
$$f\left(\frac{1}{xy + x + y}\right) f\left(\frac{1}{z}\right) = f\left(\frac{1}{xz + x + z}\right) f\left(\frac{1}{y}\right).$$
Replacing $(y, z)$ by $(y^{-1}, z^{-1})$ in above we have
$$\frac{f\left(\frac{y}{xy+x+1}\right)}{f(y)} = \frac{f\left(\frac{z}{xz+x+1}\right)}{f(z)}.$$
Since both $y$ and $z$ are arbitrary,
$$\begin{aligned} \frac{f\left(\frac{y}{xy+x+1}\right)}{f(y)} &= \frac{1}{f(y)f\left(\frac{xy+x+1}{y} + 1\right)} - \frac{1}{f(y)} && (\text{set } w = \frac{xy+x+1}{y} \text{ in (1)}) \\ &= \frac{1}{1 - f\left(\frac{1}{x+1} + \frac{1}{y}\right)} - \frac{1}{f(y)} && (\text{by } P(y^{-1}, \frac{y}{(x+1)(y+1)})) \end{aligned}$$
is constant in $y$ as $x$ fixed. Replace $x$ by $x-1$ (for $x > 1$) in above, then
$$\frac{1}{1 - f\left(\frac{1}{x} + \frac{1}{y}\right)} - \frac{1}{f(y)} - \frac{1}{f(x)}$$
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## 2024-TWN — Page 81
is independent of $y$. By symmetry, it is also independent of $x$. Say
$$Q(x, y) : \quad \frac{1}{1 - f\left(\frac{1}{x} + \frac{1}{y}\right)} - \frac{1}{f(y)} - \frac{1}{f(x)} = c, \quad \forall x, y > 1,$$
for some constant $c \in \mathbb{R}$. For $x, y, z > 2$, we have $\frac{xy}{x+y}, \frac{xz}{x+z} > 1$. Comparing $Q(\frac{xy}{x+y}, z)$ and $Q(\frac{xz}{x+z}, y)$, we have
$$\frac{1}{f\left(\frac{xz}{x+z}\right)} + \frac{1}{f(y)} = \frac{1}{1 - f\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)} - c = \frac{1}{f\left(\frac{xy}{x+y}\right)} + \frac{1}{f(z)}.$$
Hence
$$\frac{1}{f\left(\frac{xy}{x+y}\right)} - \frac{1}{f(y)} - \frac{1}{f(x)}$$
is constant in $y > 2$ as $x$ fixed. By symmetry, it is also independent of $x$, and let the constant be $c'$. Define
$$g(t) = \frac{1}{f(t^{-1})} + c'$$
on $\mathbb{R}_+$. Then for $x, y > 2$,
$$g(x^{-1} + y^{-1}) = \frac{1}{f\left(\frac{xy}{x+y}\right)} + c' = \left( \frac{1}{f(y)} + \frac{1}{f(x)} + c' \right) + c' = g(x^{-1}) + g(y^{-1}).$$
So $g$ is a Cauchy function on $(0, 0.5)$. Since $g$ has lower bound $c'$, $g(x) = ax$ on $(0, 0.5)$. By the above equation, $g(x) = ax$ on $(0, 1)$. Hence
$$f(t^{-1}) = \frac{1}{at - c'}$$
for $t \in (0, 1)$, i.e.,
$$f(x) = \frac{x}{a - c'x}, \quad x > 1.$$
Since $w + 1 > 1$ for $w > 0$ and by (1), we have
$$f(w^{-1}) = f(w + 1)^{-1} - 1 = \frac{a}{w + 1} - (c' + 1), \quad \forall w > 0.$$
Hence $f(x) = \frac{ax}{x+1} + b$ for some $a, b \in \mathbb{R}$. Then $b \ge 0$ by taking $x$ small enough. Substituting the original condition, we have
$$1 = f(xy + x + y) + f(x^{-1})f(y^{-1}) = \frac{a(xy + x + y)}{(x+1)(y+1)} + b + \left(\frac{a}{x+1} + b\right) \left(\frac{a}{y+1} + b\right),$$
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## 2024-TWN — Page 82
i.e.,
$$xy + (x + y) + 1 = (a + b + b^2)xy + (a + b + ab + b^2)(x + y) + b + (a + b)^2.$$
Comparing the coefficients in the above equation, we have
$$a + b + b^2 = 1 = a + b + ab + b^2 \implies ab = 0.$$
If $a = 0$, then $b + b^2 = 1$, i.e., $b = \frac{-1+\sqrt{5}}{2}$ and $f(x) = b$. If $b = 0$, then $a = 1$ and thus $f(x) = \frac{x}{x+1}$. $\square$

Remark. There is another short solution as follows. Notice that $\text{Im } f \subset (0, 1)$ and consider $g : (0, 1) \to (0, 1)$ defined by
$$g(x) = f\left(\frac{x}{1-x}\right).$$
Then
$$P\left(\frac{1-x}{x}, \frac{1-y}{y}\right) : \quad 1 = f\left(\frac{1-xy}{xy}\right) + g(x)g(y) = g(1-xy) + g(x)g(y), \quad \forall x, y \in (0, 1),$$
and denoted it by $R(x, y)$. Then $R(xy, z)$ and $R(x, yz)$ show that
$$g(xy)g(z) = g(x)g(yz) \implies \frac{g(xy)}{g(x)g(y)} = \frac{g(xz)}{g(x)g(z)}.$$
By the same argument above, $\frac{g(xy)}{g(x)g(y)}$ is independent with respect to $x$ and $y$, so it must be a constant $c > 0$. Define $h : (-\infty, 0) \to (-\infty, \log c)$ by
$$h(x) = \log g(e^x) + \log c.$$
Then
$$h(xy) = \log(cg(e^{xy})) = \log(c^2g(e^x)g(e^y)) = h(x) + h(y),$$
i.e., $h$ is a Cauchy function having an upper bound, so it must be linear on $(0, 1)$. Hence we can solve that $f$ is also linear in $\mathbb{R}_+$.