Anyone ?

$y=1: f(x+f(x))=xf(1+f(1))$ Set $f(1+f(1))=C$ $f(x+f(x))=Cx$ $x=f(t)+t,y=1: f(t+f(t)+f(t+f(t)))=f(t+f(t)+Ct)=C(t+f(t))$ $x=(C+1)t,y=\frac{1}{C+1}$ : $f((C+1)t+f(t))=(C+1)t f(1+f(\frac{1}{C+1}))=C(t+f(t))$ $\frac{(C+1)f(1+f(\frac{1}{C+1}))}{C}-1=D$ $f(t)=Dt$ $xf(1+f(y))=Dx+D^2xy$ $f(x+f(xy))=f(x+Dxy)=Dx+D^2xy$ So $f(x)=Dx$ is solution.

is 0 included or not ??

No, $0\notin\mathbb R^+$.

Here is my solution. Let $x = t - f(0)$ and $y = 0$. Then we have $$f(t) = (t - f(0))f(1 + f(0))$$from where we obtain that $f(t) = at + b$ where $a = f(1 + f(0))$ and $b = - f(0)f(1 + f(0))$ so our function is linear and plugging this back in equation gives us $(a +1)b = 0$ so $f(t) = at$ or $f(t) = b - t$. Hence we are done.

AleksaS wrote: Here is my solution. Let $x = t - f(0)$ and $y = 0$. Then we have $$f(t) = (t - f(0))f(1 + f(0))$$from where we obtain that $f(t) = at + b$ where $a = f(1 + f(0))$ and $b = - f(0)f(1 + f(0))$ so our function is linear and plugging this back in equation gives us $(a +1)b = 0$ so $f(t) = at$ or $f(t) = b - t$. Hence we are done. Well... math90 wrote: No, $0\notin\mathbb R^+$. So $f(0)$ doesnt exist.

Yup, thanks for warning, I obtained same solution as RagvaloD.

I get f(x)=x+1?

FriIzi wrote: I get f(x)=x+1? $f(x)=x+1$ is not a solution

Let $P(x,y)$ denote the assertion. Lemma1: f is non decreasing Proof: $P(x,1) \implies f(x+f(x))=xf(1+f(1))$ so, $$f(p+q+f((p+q)x))=f(p+f(px))+f(q+f(qx)) [Q(x,p,q)]$$for all $x,p,q \in \mathbb{R}^+$. Writing $f(x)=f(x+y)+z$ for some $x,y,z \in \mathbb{R}^+$ contradicts with $Q(\frac{y}{z},\frac{xz}{y},z)$. Lemma2: f is surjective Lemma3: f is bijective Proof: Suppose $f(\alpha)=f(\beta)$ for some $\alpha > \beta$. Then $P(x,\alpha), P(x,\beta)$ gives us that f is constant in the interval $[x\alpha, x\beta]$ by the non decreasing property. So f is constant everywhere, a contradiction. Lemma4: f is increasing and continious. Let $f(1/c)=1$. $$P(\frac{1}{f(1+f(y))},y) \implies f(1+f(y)) > c \implies f(x+f(xy)) > cx$$Since $P(\frac{1}{N}, y)$ means that we can find arbitrarily close values to zero, we conclude that $f(y) > x$ for all $y \in (x, \infty)$, so by continuity $f(x) \geq cx$. Suppose $f(x)=cx + \epsilon$ for some $x$. $P(\frac{f(x)}{f(1+f(y))}, y) \implies c+\frac{\epsilon}{x} \leq f(1+f(y))$ A contradiction again since $f(y)$ can have an arbitrarily small value again. So $f(x)=cx$.

ZETA_in_olympiad wrote: FriIzi wrote: I get f(x)=x+1? $f(x)=x+1$ is not a solution .....