# Find all functions such that $f(1+xf(y))=yf(x+y)$ where $x,y \in R^+$

Find all functions run over positive real numbers such that $f(1+xf(y))=yf(x+y)$ where $x,y\in R^+$

MY ANSWER: Putting $x=y=0$,we get, $f(1)=0$ Putting $x=0$ we get, $f(1)=yf(y)$

or,$yf(y)=0$

or,$f(y)=0$ (since $y\ne 0$., $y \in \mathbb R^+$)

Hence,$f(x)=0$ is the solution.

Is my answer and solution collect? If not then please tell me the proper answer and solution and where I have made the mistake!!

• The $x=y=0$ does not seem valid, since on right side you have $f(0)$ and $0 \not\in \mathbb{R}^+$. Or what is domain and codomain of the function? – Sil Jun 17 '18 at 14:13
• Your solution does not seem to apply to $f(0)$. If I may, an alternative approach could be using $y=1$ which gives $f(1)=f(x+1)\ \forall x\in\mathbb{R^+}\Rightarrow f(x)=0\ \forall x\in\mathbb{R^+}$, using that $f(1)=0$ – user418131 Jun 17 '18 at 14:15
• Btw here this is discussed on AoPS, there should be some solutions: artofproblemsolving.com/community/c6h323174 . – Sil Jun 17 '18 at 14:18
• I am satisfied. Thank you! – Sufaid Saleel Jun 17 '18 at 14:27

Replacing $$x$$ by $$\frac{x}{f(y)}$$ in $$(1),$$ we have $$f(1+x)=y\cdot f\left(\frac{x}{f(y)}+y\right),\quad \forall x ,\,y \in \mathbb R^+. \quad (2)$$ Next, we replace $$y$$ by $$\frac{f(1+x)}{y}$$ in $$(2)$$ to get $$y=f\left(\frac{x}{f\left(\frac{f(1+x)}{y}\right)}+\frac{f(1+x)}{y}\right),\quad \forall x,\,y \in \mathbb R^+. \quad (3)$$ From $$(3),$$ it follows that $$f$$ is surjective. Now, we will prove that $$f$$ is decreasing. Replacing $$x$$ by $$x+z$$ in $$(1),$$ we get $$f\big(1+(x+z)\cdot f(y)\big)=y\cdot f(x+y+z),\quad \forall x,\,y,\,z \in \mathbb R^+. \quad (4)$$ Replacing $$y$$ by $$y+z$$ in $$(1),$$ we also have $$f\big(1+x\cdot f(y+z)\big)=(y+z)\cdot f(x+y+z),\quad \forall x,\,y,\,z \in \mathbb R^+. \quad (5)$$ Dividing $$(4)$$ for $$(5),$$ side by side, we obtain $$\frac{f\big(1+(x+z)\cdot f(y)\big)}{f\big(1+x\cdot f(y+z)\big)}=\frac{y}{y+z},\quad \forall x,\,y,\,z \in \mathbb R^+. \quad (6)$$ Now, assume that there exists a pair $$(y,\,z)$$ such that $$f(y+z)>f(y).$$ In this case, by choosing $$x=\frac{z\cdot f(y)}{f(y+z)-f(y)}$$ vào $$(6),$$ we obtain $$y=y+z,$$ which is a contradiction. So we must have $$f(y+z) \le f(y),\quad \forall y,\,z \in \mathbb R^+. \quad (7)$$ Now, we will prove that $$f$$ is injective. Assume that there are two numbers $$a,\,b$$ such that $$f(a)=f(b).$$ Replacing $$y=a$$ and $$y=b$$ in $$(1)$$ respectively in $$(1),$$ we get $$a\cdot f(x+a)=b\cdot f(x+b),\quad \forall x\in \mathbb R^+. \quad (8)$$ From this, it follows that $$1+a(y-1)\cdot f(x+a)=1+b(y-1)\cdot f(x+b),\quad \forall x, \, y \in \mathbb R^+,\, y>1. \quad (9)$$ Plugging this into $$f$$ and using $$(1),$$ we get $$(x+a)\cdot f(x+ay)=(x+b)\cdot f(x+by),\quad \forall x,\,y \in \mathbb R^+ ,\, y>1. \quad (10)$$ From $$(10),$$ it follows that, for any $$x,\,y,\,z \in \mathbb R^+,\, y>1,$$ $$1+(xz+az)\cdot f(x+ay)=1+(xz+bz)\cdot f(x+by). \quad (11)$$ Again, we plug this into $$f$$ and using $$(1).$$ It follows that $$(x+ay)\cdot f(x+ay+az+xz)=(x+by)\cdot f(x+by+bz+xz)\quad (12)$$ for any $$x,\,y,\,z \in \mathbb R^+$$ and $$y>1.$$ On the other hand, according to $$(10),$$ we also have $$\big[(x+xz)+a\big]\cdot f\big( (x+xz)+a(y+z)\big)=\big[(x+xz)+b\big] \cdot f\big((x+xz)+b(y+z)\big),$$ or $$(x+xz+a)\cdot f(x+ay+az+xz)=(x+xz+b)\cdot f(x+ay+az+xz). \quad (13)$$ Dividing $$(12)$$ for $$(13),$$ side by side, we obtain $$\frac{x+ay}{x+xz+a}=\frac{x+by}{x+xz+b},\quad \forall x,\,y,\,z \in \mathbb R^+,\, y>1. \quad (14)$$ It is easy to deduce that $$a=b$$ here, so $$f$$ is injective. Now, replacing $$x=y=1$$ in $$(1)$$ with notice that $$f$$ is injective, we have $$f(1)=1.$$ Since $$f$$ is strictly decreasing ($$f$$ is decreasing and injective), we have $$f(x)<1,\quad \forall x>1.\quad (15)$$ Now, let us consider the case $$y>x.$$ Replacing $$y$$ by $$y-x$$ in $$(1)$$ and using the above remark, we get $$f(y)=\frac{f\big(1+x\cdot f(y-x)\big)}{y-x}<\frac{1}{y-x},\quad \forall x,\,y \in \mathbb R^+,\, x In $$(16),$$ we let $$x\to 0^+$$ and obtain $$f(y) \le \frac{1}{y},\quad \forall y >0. \quad (17)$$ Next, replacing $$x$$ by $$x-1$$ and $$y$$ by $$f(y)$$ in $$(3),$$ we get $$y=\frac{x-1}{f\left(\frac{f(x)}{f(y)}\right)}+\frac{f(x)}{f(y)},\quad \forall x ,\, y \in \mathbb R^+,\, x >1. \quad (18)$$ Since $$f\left(\frac{f(x)}{f(y)}\right) \le \frac{f(y)}{f(x)},$$ from $$(18),$$ we deduce that $$y\ge \frac{(x-1)\cdot f(x)}{f(y)}+\frac{f(x)}{f(y)},$$ or $$y\cdot f(y) \ge x \cdot f(x),\quad \forall x,\,y \in \mathbb R^+,\, x>1. \quad (19)$$ Changin the position of $$x$$ and $$y$$ in $$(19),$$ we also have $$x\cdot f(x) \ge y\cdot f(y),\quad \forall x,\,y \in \mathbb R^+,\, y>1. \quad (20)$$ From the inequalities $$(19)$$ and $$(20),$$ we can easily deduce that $$f(x)=\frac{k}{x},\quad \forall x>1. \quad (21)$$ Now, taking $$x=1$$ in $$(1)$$ and using $$(21),$$ we have $$\frac{1}{1+f(y)}=\frac{y}{1+y},$$ or $$f(y)=\frac{1}{y},\quad \forall y\in \mathbb R^+. \quad (22)$$ Clearly, the function $$f(x)=\frac{1}{x}$$ satisfies our equation.