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Find all functions $f\colon \mathbb{R}^* \to \mathbb{R}^* $ from the non-zero reals to the non-zero reals, such that $$f(xyz)=f(xy+yz+xz)(f(x)+f(y)+f(z))$$ for all non-zero reals $x, y, z$ such that $xy+yz+xz\neq 0$.

I think that only two solutions are: $f(x)=\frac{1}{3}$ and $f(x)=\frac{1}{x}$.

I would appreciate any suggestions.

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  • $\begingroup$ $f(x) = \frac 13$ is not an injective function. But, I haven't thought of any others, either. $\endgroup$ – Doug M Jun 23 '16 at 2:28
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Plugging in $(x,y,z)=(1,1,-1)$ and $(x,y,z)=(1,-1,-1)$, we get $$\begin{eqnarray*} f(-1)&=&f(-1)(2f(1)+f(-1))\\ f(1)&=&f(-1)(2f(-1)+f(1)) \end{eqnarray*}$$ Solving, $f(1)=1$ or $\frac13$.

Case 1: $f(1)=1$

For any $a\neq0$, plugging in $(x,y,z)=(a,-a,1)$ gives $$ f(-a^2)=f(-a^2)(f(a)+f(-a)+1). $$ Hence $f(-a)=-f(a)$. For any $a,b>0$, putting $(x,y,z)=(\sqrt{a},-\sqrt{a},b)$ gives $$ f(-ab)=f(-a)f(b) $$ Hence $f(ab)=f(a)f(b)$.

Let $g(x)=\frac1{f(x)}$, so $g(ab)=g(a)g(b)$ for $a,b>0$. In particular if $a>0$ then $g(a)=g(\sqrt{a})^2>0$. The original equation becomes $$ g(xy+yz+zx)=g(xyz)\left(\frac1{g(x)}+\frac1{g(y)}+\frac1{g(z)}\right) =g(xy)+g(yz)+g(zx). $$ For any $a,b,c>0$, setting $(x,y,z)=(\sqrt{ac/b},\sqrt{ab/c},\sqrt{bc/a})$ gives $$ g(a+b+c)=g(a)+g(b)+g(c). $$ In particular if $a>b>0$ then $g(a)=g(b)+2g(\frac{a-b}2)>g(b)$. Note that $g(3)=3g(1)=3$, so $g(3^k)=3^k$ for all integers $k$. Thus for any $a,b>0$, $$ g(a+b)\leq g\left(a+b+3^k\right)=g(a)+g(b)+3^k. $$ Sending $k\rightarrow-\infty$, $g(a+b)\leq g(a)+g(b)$. On the other hand, pick $k$ such that $3^k<b$. Then $$ g(a)+g(b)\leq g(a)+g(b-3^k)+g(3^k)=g(a+b). $$ Hence $g(a+b)=g(a)+g(b)$. Since $g(1)=1$, $g(x)=x$ on positive rationals. Also $g$ is increasing, so $g(x)=x$ on positive reals. Since $f$ is odd, $f(x)=\frac1x$ for all $x\in\mathbb R^*$.

Case 2: $f(1)=\frac13$

For any $a\neq0$, plugging in $(x,y,z)=(a,-a,1)$ gives $$ f(-a^2)=f(-a^2)(f(a)+f(-a)+\frac13). $$ Hence $f(a)+f(-a)=\frac23$. Putting $(x,y,z)=(a,-a,b^2)$ gives $$ f(-a^2b^2)=f(-a^2)(\frac23+f(b^2))=f(-a^2)(\frac43-f(-b^2)). $$ Thus $$ f(-a^2b^2)+f(-a^2)f(-b^2)=\frac43f(-a^2). $$ Swapping $a$ and $b$, $f(-a^2)=f(-b^2)$. In particular, putting $b=1$, $f(-a^2)=\frac13$. Hence $f(a)=\frac13$ for $a<0$, and for $a>0$ we have $f(a)=\frac23-f(-a)=\frac13$ also.

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  • $\begingroup$ Nice solution! Thank you! $\endgroup$ – MoM-ADMIN Jun 23 '16 at 10:13
  • $\begingroup$ How can you get conclusion "g(x) = x on positive real" from "g(x) = x on positive rationals"? $\endgroup$ – Zack Ni Jun 23 '16 at 12:09
  • $\begingroup$ @ Zack Ni, Because $g$ is increasing and $\mathbb{Q}$ is dense in $\mathbb{R}.$ $\endgroup$ – MoM-ADMIN Jun 23 '16 at 14:01

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