Iranian Mathematical Olympiad

Let $P(x,y)$ be the assertion $f\left(\frac{y}{f(x+1)}\right)+f\left(\frac{x+1}{xf(y)}\right)=f(y)$.

Rewriting the equation, we get $f\left(\frac{y}{f(x)}\right)+f\left(\frac{x}{(x-1)f(y)}\right)=f(y)$ which is valid for each $x>1,y\in {\mathbb{R}}^{+}$.

If there exists $a\in {\mathbb{R}}^{+}$ for which $f(a)>\frac{1}{a}$, we get $$P\left(\frac{af(a)}{af(a)-1},a\right)\Rightarrow f\left(\frac{a}{f\left(\frac{af(a)}{af(a)-1}\right)}\right)=0,$$ which is a contradiction. Therefore, for each $x\in {\mathbb{R}}^{+}$, $f(x)\le \frac{1}{x}$.

Now, we can write $$P(x>1,y)\Rightarrow f(y)=f\left(\frac{y}{f(x)}\right)+f\left(\frac{x}{(x-1)f(y)}\right)\le \frac{f(x)}{y}+\left(1-\frac{1}{x}\right)f(y)\Rightarrow yf(y)\le xf(x).$$ So for all $x,y>1\in {\mathbb{R}}^{+}$, $xf(x)=yf(y)$. Hence for each $x>1\in {\mathbb{R}}^{+}$, $f(x)=\frac{C}{x}$ where $C$ is a constant number. If we choose real numbers $x,y$ greater than 1 such that $y$ is also greater than $C$, substituting these values for $x$ and $y$ in the equation shows that $C=1$. Therefore, $\forall x>1\in {\mathbb{R}}^{+}$, $f(x)=\frac{1}{x}$.

Since $f(1)\le 1$, we have $$\begin{array}{rl}P(x>1,1)& \Rightarrow f(1)=f\left(\frac{1}{f(x)}\right)+f\left(\frac{x}{(x-1)f(1)}\right)=f(x)+\frac{x-1}{x}f(1)\\ & \Rightarrow \frac{1}{x}f(1)=\frac{1}{x}\Rightarrow f(1)=1.\end{array}$$ Now, for $\frac{1}{2}\le x<1$, we can write $$\begin{array}{rl}P(2,x)& \Rightarrow f\left(\frac{x}{f(2)}\right)+f\left(\frac{2}{f(x)}\right)=f(x)\Rightarrow f(2x)+f\left(\frac{2}{f(x)}\right)=f(x)\\ & \Rightarrow \frac{1}{2x}+\frac{f(x)}{2}=f(x)\Rightarrow f(x)=\frac{1}{x}.\end{array}$$ In a similar way, by using induction on $n$ one can prove that for each positive real $x$ in the interval $\frac{1}{2^n}\le x<\frac{1}{2^{n-1}}$, $f(x)=\frac{1}{x}$. Therefore, for each $x\in {\mathbb{R}}^{+}$, $f(x)=\frac{1}{x}$. It's easy to verify that this solution is indeed an answer to the functional equation.