Find all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for all $x,y \in \mathbb{R}-\{0\}$, $$ f(x) \neq 0 \text{ and } \frac{f(x)}{f(y)} + \frac{f(y)}{f(x)} - f \left( \frac{x}{y}-\frac{y}{x} \right) =2 $$
Problem
Source: 2025 Turkey TST P3
Tags: functional equation, Turkey, algebra
swynca
18.03.2025 11:27
There is a single solution for $f$ which is $f(x)=x^2$.
Let $P(x,y)$ refer to the equation $\frac{f(x)}{f(y)} + \frac{f(y)}{f(x)} - f \left( \frac{x}{y}-\frac{y}{x} \right) =2 $,
$P(x,x) \rightarrow f(0)=0$
$P(x,-x) \rightarrow $ after using AGM, $f(x)=f(-x)$ $\textbf{(I)}$
We know that $f(\alpha)=0 \iff \alpha=0$. Using that if $f(\alpha)=f(\beta)$,
$P(\alpha,\beta) \rightarrow$ $f \left( \frac{\alpha}{\beta}-\frac{\beta}{\alpha} \right) =0$ which means $\alpha^2=\beta^2$. So combining with $\textbf{(I)}, $we have $f(x)=f(y) \iff x^2=y^2$.
If we use the solutions of the equation $\frac{x}{y}-\frac{y}{x}= \frac{x}{z}-\frac{z}{x}$,
for $x^2\neq y^2$ and $x,y \neq 0$, $P(x,y)$ and $P(x,\frac{x^2}{y})$ gives us, $f(x)^2=f(\frac{x^2}{y})f(y)$
which also means, for $x^2 \neq 1$ and $x,y \neq0$, $f(xy)^2=f(x^2)f(y^2)$. Let $Q(x,y)$ refer to this equation. It shows that for all $x \neq 0$, $f(x) > 0$ or $f(x) <0$. If $f(x)<0$ holds, then using AGM gives us a contradiction on $P(x,y)$. Thus we have $f(x)\geq0$ for all $x$.
$Q(x,1) \rightarrow f(x)^2=f(x^2)f(1)$. Both equations hold for $x^2=1$, and also for $x=0$ and $y=0$. So we don't need any limitations for $Q(x,y)$.
Let's say $f(1)=c$. We know that $f(x)^2=f(x^2)c$. Puting it on $Q(x,y)$ gives us $f(xy)c=f(x)f(y)$.
Applying on $P(x,y)$ implies,
$(f(x)-f(y))^2 = f(\frac{x^2-y^2}{xy})f(x)f(y)=f(x^2-y^2)c^2$ $\textbf{II}$
For $x^2 \neq y^2$, $P(x,\sqrt{\frac{x^2+y^2}{2}})$ and $P(y,\sqrt{\frac{x^2+y^2}{2}})$ shows that, $f(x)+f(y)= 2f(\sqrt{\frac{x^2+y^2}{2}})$. $\textbf{(III)}$
Thus, for all $0<a<2$, $f(a)+f(\sqrt{2-a})=2c$. Which implies $f(a)<2c$.
Finally we will apply a functional transformation to convert multiplicativeness into additivity. Let's define a function $h: \mathbb{R}\rightarrow \mathbb{R}$ such that, $h(x)=ln \left( \frac{f(e^x)}{c} \right)$. It can also shown as for $x>0$, $f(x)= e^{h(lnx)}c$. Implementing in $Q(x,y)$ and doing some manipulation shows us:
$$h(x)+h(y)=2(\frac{x+y}{2})$$This is a jensen function which can easily be converted into cauchy functional equation. By definition, we know that $h(0)=0$. So, $h(x)=2h(\frac{x}{2})$.
$$\Rightarrow h(x)+h(y)=2h(\frac{x+y}{2})=h(x+y)$$We proved that for $0<a<2$, $0<f(a)<2c$. Which implies that, for $\alpha<ln2$ we have $h(\alpha)<ln2$. Thus $h$ is bounded in an interval. So, $h$ is linear. $h(0)=0$.
$$\Rightarrow h(x)=dx$$for some constant $d$. Now, we have, $f(x)=x^dc$. Using in $\textbf{III}$ gives us,
$$\left(\frac{x^d+y^d}{2}\right)^{\frac{1}{d}}=\left( \frac{x^2+y^2}{2} \right)^{\frac{1}{2}}$$Generalized Mean Inequality shows that equality can't hold since $x\neq y$, for any $d \neq 2$. So, $d=2$. One can easily check that for $f(x)=x^2c$, the only $c$ that satisfies the condition is $c=1$. $\blacksquare$
AnSoLiN
18.03.2025 14:44
let $P(x,y)$ be the given equation. Claim 1. $f(x)=0\iff x=0$ Proof. By $P(x,x)$ we get $f(0)=0$. And for any $x\neq 0$, we have $f(x)\neq 0$ by the problem statement. Claim 2. $f(x)=f(y)\iff x^2=y^2$ Proof. If $f(x)=f(y)$, we get $f\left(\frac{x}{y}-\frac{y}{x}\right)=0$ and by Claim 1, we get $\frac{x}{y}-\frac{y}{x}=0$, which means $x^2=y^2$. On the other hand, by $P(x,y)$ and $P(y,x)$, we get $f(x)=f(-x)$. Lemma. $$a+\frac{1}{a}=b+\frac{1}{b}\implies (a-b)\left(1-\frac{1}{ab}\right)=0\implies a=b\text{ or }a=\frac{1}{b}$$ Claim 3. There is no $x,y$ such that $f(y)<0<f(x)$ Proof. Partition the set $\mathbb{R}-\{0\}$ into two groups $A=\{x:f(x)>0\}$ and $B=\{x:f(x)<0\}$. Let $x\sim y$ denote $x$ and $y$ are in the same group. If $x\sim y$, we get $f\left(\frac{x}{y}-\frac{y}{x}\right)\geq 0$ because $\frac{f(x)}{f(y)}+\frac{f(y)}{f(x)}\geq 2$, and on the other hand if $x\not\sim y$, we get $f\left(\frac{x}{y}-\frac{y}{x}\right)\leq -4$ because $\frac{f(x)}{f(y)}+\frac{f(y)}{f(x)}\leq -2$. These observations mean $x\sim y\implies xz\sim yz$. Then, for any $u\neq 0$, depending on whether or not $1\sim u$, we have either $x\sim xu\sim xu^2$ or $x\not\sim xu\not\sim xu^2$, both of which give $x\sim xu^2$. That is, multiplying $x$ with a positive number does not change its group. With Claim 2, we get either $A$ or $B$ is empty. Claim 4. For all $x,y\in\mathbb{R}$, we have $f(xy)f(1)=f(x)f(y)$ Proof. If $x=0$ or $y=0$, it is true. So let's assume $x,y\in\mathbb{R}-\{0\}$. By $P(1,y)$ and $P(x,xy)$, we get $$\frac{f(1)}{f(y)}+\frac{f(y)}{f(1)}-f\left(\frac{1}{y}-y\right)=2=\frac{f(x)}{f(xy)}+\frac{f(xy)}{f(x)}-f\left(\frac{1}{y}-y\right)$$With the lemma, this means $$f(xy)f(1)=f(x)f(y)\text{ or }f(xy)f(y)=f(x)f(1)$$Assume there exists $x,y$ such that $f(xy)f(1)\neq f(x)f(y)$. Then we must have $f(xy)f(y)=f(x)f(1)$ and $f(xy)f(x)=f(y)f(1)$. Multiplying these, we get $f(x)^2=f(y)^2$ and $f(xy)^2=f(1)^2$. By Claim 3, we get $f(x)=f(y)$ and $f(xy)=f(1)$, and by Claim 2, we get $x^2=y^2$ and $(xy)^2=1$. That means we have cases $(x,y)=(1,1),(1,-1),(-1,1),(-1,-1)$, all of which already satisfies $f(xy)f(1)=f(x)f(y)$, hence the result follows. Now that we get a nice multiplicative property, we will use it in the main equation. For all $x,y\in\mathbb{R}-\{0\}$, $$f\left(\frac{x}{y}-\frac{y}{x}\right)=f\left(\frac{x^2-y^2}{xy}\right)=\frac{f(x^2-y^2)f(1)}{f(xy)}=\frac{f(\sqrt{|x^2-y^2|})^2f(1)}{f(x)f(y)}$$so $$\frac{f(x)}{f(y)}+\frac{f(y)}{f(x)}-\frac{f(\sqrt{|x^2-y^2|})^2f(1)}{f(x)f(y)}=2\implies f(x)^2+f(y)^2-2f(x)f(y)-f(1)f(\sqrt{|x^2-y^2|})^2=0$$Let $R(x,y)$ be the last equation. Let's look at $R(x,z)$ and $R(y,z)$ where $z=\sqrt{x^2+y^2}$. $$f(x)^2+f(z)^2-2f(x)f(z)-f(1)f(y)^2=0$$$$f(y)^2+f(z)^2-2f(y)f(z)-f(1)f(x)^2=0$$Subtracting one from other, we get $$(f(1)+1)(f(x)-f(y))(f(x)+f(y))=2f(z)(f(x)-f(y))$$If $x^2\neq y^2$, then by Claim 2, we have $f(x)\neq f(y)$, so we get $f(z)=k(f(x)+f(y))$ where $k=\frac{f(1)+1}{2}$. Rewrite this in $R(x,z)$ (or $R(y,z)$) and we get $$(k^2-f(1))f(x)^2+(k^2-f(1))f(y)^2+(2k^2-f(1)-1)f(x)f(y)=0$$If $f(1)\neq 1$, then $k^2\neq f(1)$ and take any different $x_1,x_2,x_3>0$. $x_1,x_2,x_3$ all satisfy this quadratic equation if we fix a $y\neq 0$, which is not possible. So $f(1)=1=k$ and $f(z)=f(x)+f(y)$. We get a nice additive property, time to combine this with multiplicativity. Let $g(x)=f(\sqrt{|x|})$. Then we have $g(x+y)=g(x)+g(y)$ and $g(xy)g(1)=g(x)g(y)$, which means $g(x)=g(1)x$ and $f(x)=cx^2$. Putting this into main equation, we get $c=1$ and the only solution $\boxed{f(x)=x^2}$