let $p(x,y)= f(\frac{y}{f(x+1)})+f(\frac{x+1}{xf(y)})=f(y)$ if we have $y$ such that $yf(y)> 1$,$x=\frac{1}{yf(y)-1}\Rightarrow f(\frac{y}{f(x+1)})=0$ which is not true. so for all $x$ we have $f(x)\leq \frac{1}{x}$ . $ f(y)=\frac{y}{f(x+1)})+f(\frac{x+1}{xf(y)})\leq \frac{x}{x+1}f(y)+f(\frac{y}{f(x+1)})\Rightarrow \frac{f(y)}{x+1}\leq f(\frac{y}{f(x+1)})\leq f(x+1).\frac{1}{y}\Rightarrow yf(y)\leq (x+1)f(x+1)\Rightarrow $ for all $1<x$ $f(x)=\frac{a}{x}$ $\rightarrow yf(y)\leq a\leq 1$ $p(x,y+1)\Rightarrow a=1$ $p(x,1)\Rightarrow f(1)=1$ $p(x,\frac{1}{x+1}),(\frac{x+1}{xf(\frac{1}{x+1})}> 1)\Rightarrow f(\frac{1}{x+1})=x+1\Rightarrow f(x)=\frac{1}{x}$ so we are done

i think it is not good problem for TST . similar idea : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=578815&p=3415155#p3415155 and this : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=576480&p=3397645#p3397645

nima1376 wrote: ...$\Rightarrow f(\frac{1}{x+1})=x+1\Rightarrow f(x)=\frac{1}{x}$ so we are done I've not read the entire proof but the line above is wrong. The good implication would be $f(\frac{1}{x+1})=x+1\Rightarrow f(x)=\frac{1}{x}$ $\boxed{\forall x>1}$ And so it remains to find the values for $x\in(0,1]$

i proved $ f(x)=\frac{1}{x} $ for $x>1$ then i proved $\Rightarrow f(\frac{1}{x+1})=x+1$ and then i prove $f(x)=\frac{1}{x}$

nima1376 wrote: ...and then i prove $f(x)=\frac{1}{x}$ I did not see where. If you say it, it's certainlay true ...

nima1376 wrote: for all $1<x$ $f(x)=\frac{a}{x}$ $p(x,y+1)\Rightarrow a=1$

nima1376 wrote: $p(x,\frac{1}{x+1}),(\frac{x+1}{xf(\frac{1}{x+1})}> 1)\Rightarrow f(\frac{1}{x+1})=x+1\Rightarrow f(x)=\frac{1}{x}$ How is $\frac{x+1}{xf(\frac{1}{x+1})} > 1$? We would require $f(\frac{1}{x+1}) < 1+\frac{1}{x}$ but that's obviously false. Here's a solution. Denote $P(x,y)$ what it usually means. Step 1. Assume that there exists an $y$ such that $yf(y)>1$. Now do a $P(\frac{1}{yf(y)-1},y)$. This gives a contradiction, since this forces $f(\frac{y}{f(\frac{1}{yf(y)-1}+1)}) = 0$. Therefore, $yf(y) \le 1$ for all $y$. This gives us $f(\frac{x+1}{xf(y)}) \le \frac{xf(y)}{x+1}$, so $\frac{f(x+1)}{y} \ge f(\frac{y}{f(x+1)}) \ge \frac{f(y)}{x+1}$. Therefore, $(x+1)f(x+1) \ge yf(y)$, which gives us that $xf(x)=C \le 1$ for all $x>1$. Step 2. We prove $C=1$. $P(x,y+1)$ gives you $f(\frac{y+1}{f(x+1)}) + f(\frac{x+1}{xf(y+1)}) = f(y+1)$. This gives us $f(\frac{(x+1)(y+1)}{C}) + f(\frac{(x+1)(y+1)}{Cx}) = \frac{C}{y+1}$ Now since $C \le 1$, we have $\frac{(x+1)(y+1)}{C} \ge 1$ and $\frac{(x+1)(y+1)}{Cx} \ge 1$. This gives us $\frac{C^2}{(x+1)(y+1)} + \frac{C^2x}{(x+1)(y+1)} = \frac{C}{y+1}$, which gives us $C^2=C$, or $C=1$. Therefore, for $x>1$, we have $f(x)=\frac{1}{x}$. $P(x,1)$ gives you $f(\frac{1}{f(x+1)})+f(\frac{x+1}{xf(1)})=f(1)$. Since $f(1) \le 1$, we have $\frac{x+1}{xf(1)} \ge 1$. This gives us $\frac{1}{x+1} + \frac{xf(1)}{x+1} = f(1)$, so $f(1)=1$. This gives us $f(x)=\frac{1}{x}$ for $x \ge 1$. Step 3. $P(x,\frac{x}{x+1})$ gives $f(\frac{x}{(x+1)f(x+1)}) + f(\frac{x+1}{xf(\frac{x}{x+1})}) = f(\frac{x}{x+1})$. Now note that $f(\frac{x}{x+1}) \le \frac{x+1}{x}$, so $\frac{x+1}{xf(\frac{x}{x+1})} \ge 1$. This gives us $f(x)+\frac{x}{x+1}f(\frac{x}{x+1}) = f(\frac{x}{x+1})$, so $(x+1)f(x)=f(\frac{x}{x+1})$. Let $x=\frac{k}{1-k}$, and this function of $0 \le k <1$ can express all reals. This gives us $f(k)=\frac{1}{1-k}f(\frac{k}{1-k})$. Step 4. We prove that $f(x)=\frac{1}{x}$ holds for all $x<1$. To do this, we break the interval $(0,1)$ into intervals $[\frac{1}{n+1},\frac{1}{n})$ for all naturals $n$. Base Case: $n=1$. Note that for $k \in [\frac{1}{2},1)$, $\frac{k}{1-k} > 1$. This gives us $f(k)=\frac{1}{1-k}f(\frac{k}{1-k}) = \frac{1}{1-k} \cdot \frac{1-k}{k} = \frac{1}{k}$. Inductive Step: Assume the statement for $n=l-1$. Note that for $k \in [\frac{1}{l+1},\frac{1}{l})$, we have $\frac{k}{1-k} \in [\frac{1}{l},\frac{1}{l-1})$. This gives us, with the inductive hypothesis, that $f(k)=\frac{1}{1-k}f(\frac{k}{1-k}) = \frac{1}{1-k} \cdot \frac{1-k}{k} = \frac{1}{k}$. This completes the proof of the statement for $n=l$. Given any real $\epsilon < 1$, it is enclosed in some interval in a form of $[\frac{1}{n+1}, \frac{1}{n})$. Therefore, for all $x<1$, $f(x)=\frac{1}{x}$. Therefore, we conclude that $f(x)=\frac{1}{x}$. Verification is easy. GG. $\blacksquare$

Iran TST 2014 wrote: Find all functions $f:\mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$ such that $x,y\in \mathbb{R}^{+},$ \[ f\left(\frac{y}{f(x+1)}\right)+f\left(\frac{x+1}{xf(y)}\right)=f(y) \]

Nice F.E. . We claim that $f(x)=\frac{1}{x}$ is the only function that works Let $P(x,y)$ the assertion of the given F.E. Claim 1: $f(x) \le \frac{1}{x}$ Proof: Assume that there existed $x$ such that $f(x)>\frac{1}{x}$ then by $P \left(\frac{1}{xf(x)-1},x \right)$ $$f \left(\frac{x}{f \left(\frac{1}{xf(x)-1}+1 \right)} \right)=0 \; \text{contradiction!!}$$Hence $f(x) \le \frac{1}{x}$ holds for all posititve reals $x$. Claim 2: $f(x)=\frac{1}{x}$ for all $x>1$ Proof: The ineq of Claim 1 on the F.E. gives $$\frac{f(x+1)}{y}+\frac{xf(y)}{x+1} \ge f(y) \implies (x+1)f(x+1) \ge yf(y)$$And here if u replace $y$ with $t+1$ and then swich $x$ with $t$ u get that $g(x)=xf(x)$ is constant so $f(x)=\frac{c}{x}$ for all $x>1$ and by Claim 1 we have that $c \le 1$ and now replacing $y>1$ and any $x$ on the F.E. $$\frac{c^2}{y}=f \left(\frac{y(x+1)}{c} \right)+f \left(\frac{y(x+1)}{cx} \right)=\frac{c}{y} \implies c=1$$So $f(x)=\frac{1}{x}$ for all $x>1$ as desired. Claim 3: $f$ is injective on $\mathbb R^+$ Proof: Choose $f(a)=f(b)$ and choose an $x$ larger enough such that $x>\frac{1-a}{a},\frac{1-b}{b}$ then by $P(x,a)-P(x,b)$ $$\frac{1}{a(x+1)}=\frac{1}{b(x+1)} \implies a=b \implies f \; \text{injective}$$Finishing: Now on the F.E. set a positive real $x>69^{420}$ that its insanely bigggggg $$f \left(\frac{1}{f(y)} \right)=f(y) \implies \frac{1}{f(y)}=y \implies f(y)=\frac{1}{y}$$Hence $\boxed{f(x)=\frac{1}{x} \; \forall x \in \mathbb R^+}$ its the only function that works thus we are done

Let $P(x,y)$ denote the given assertion. If $f(x)>1/x$, then $P(\tfrac 1{xf(x)-1},x)$ gives contradiction. So $P(x,y+1)$ implies $\tfrac{f(x+1)}{y+1}+\tfrac{xf(y+1)}{x+1}\geqslant f(y+1).$ This is symmetric, so $f(x)=k/x$ for $x>1.$ Plugging back we have $f(x)=1/x$, which works. $P(x,1)$ implies, since $\tfrac{x+1}{xf(1)}>1$, $f(1)=1.$ Consider $1>y\geqslant 1/2$ then, $\tfrac{y}{1-y}\geqslant 1$ and by $P(\tfrac{y}{1-y}, y)$ we have $f(y)=1/y.$ When $y\geqslant 1/3$ then, $\tfrac{y}{1-y}\geqslant 1/2$, so $f(y)=1/y.$ Iterating the argument we can cover the interval $(0,1).$