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replace $y$ with the equation below and we get: \begin{align*} y= x - f(x)\\ f(x) = f(x) +3x + yf(x-f(x)) \end{align*} and we see at the end : $ \frac{-3x}{y} = f(x-f(x)) $ the left part is a negative number but as mentioned in the question the range of $f(x)$ can't be negative.

@above, You are assuming $x> f(x)$ for all $x$ but have not shown this to be the case.

parmenides51 wrote: Prove that there is no function from positive real numbers to itself, $f : (0,+\infty)\to(0,+\infty)$ such that: $f(f(x) + y) = f(x) + 3x + yf(y)$ ,for every $x,y \in (0,+\infty)$ Let $P(x,y)$ be the assertion $f(f(x)+y)=f(x)+3x+yf(y)$ Let $a=f(1)$ If $f(u)<u$ for some $u>0$, then $P(u,u-f(u))$ $\implies$ $3u+(u-f(u))f(u-f(u))=0$, impossible So $f(x)\ge x\quad\forall x>0$ Comparing $P(x,a)$ and $P(1,f(x))$, we get $f(x)+3x+af(a)=a+3+f(x)f(f(x))\ge a+3+f(x)^2$ Which is $f(x)^2-f(x)-(3x+af(a)-a-3)\le 0$ And so $f(x)\le\frac{1+\sqrt{12x+4af(a)-4a-11}}2$ Which is clearly in contradiction with $f(x)\ge x$ when $x$ is great enough Hence the claim.

parmenides51 wrote: Prove that there is no function from positive real numbers to itself, $f : (0,+\infty)\to(0,+\infty)$ such that: $f(f(x) + y) = f(x) + 3x + yf(y)$ ,for every $x,y \in (0,+\infty)$ Claim 1 : $f(x) \geq x \quad \forall x \in \mathbb{R}^{+}$ Proof : Suppose there exists $x \in \mathbb{R^+} $such that $x > f(x)$. Then $P(x, x-f(x))$ yields $(x-f(x))f(x-f(x)) = -3x$ which is impossible. So, the claim is true. $\square$. Claim 2 : $f(x) \geq 1 \quad \forall x \in \mathbb{R}^{+}$ Proof : By Claim 1, we have $f(x) + 3x + yf(y) = f(f(x) + y) \geq f(x) + y$. So, $3x + yf(y) \geq y$. By letting $x \rightarrow 0^+$, we see that $f(y) \geq 1 \quad \forall y \in \mathbb{R^+}$. $\square$ Claim 3 : There exists a $K > 0$ such that $f(f(x)) \cdot f(x) = f(x) + 3x + K$ $\quad \forall x \in \mathbb{R^+}$ Proof : $$P(x, f(y)) : f(f(x) + f(y)) = f(x) + 3x + f(y) \cdot f(f(y))$$$$P(y, f(x)) : f(f(x) + f(y)) = f(y) + 3y + f(x) \cdot f(f(y))$$From the above 2 equations we can easily establish the existence of $K$. To prove positivity, note that $3x + K = (f(f(x)) - 1) \cdot f(x)$ which implies $3x + K > 0$. Letting $x \rightarrow 0^+$, we see that $K > 0$. $\square$ Now to finish note that $$ (4 + K) \cdot f(x) \geq f(x) + 3x + K = f(x) \cdot f(f(x)) \geq f(x) \cdot f(x) \geq x \cdot f(x)$$which yields $4 + K \geq x \quad \forall x \in \mathbb{R^+}$ which doesn’t hold true for all large enough $x$. $\blacksquare$

This one was proposed by Greece, Athanasios Kontogeorgis (aka socrates)

compare P(x,f(y)+1),P(y,f(x)+1)

My solution is similar to @pco 's solution but u have a slightly different approach in the end We can easily see that $f(x)\ge x$ by assuming the opposite : $f(x)< x $ and putting $y=x-f(x)$ we get that : $ \frac{-3x}{y} = f(x-f(x)) $ $\implies f(x)$ can be negative $\implies$ contradiction $P(x,f(y))$ yields $$f(f(x)+f(y))=f(x)+3x+f(y)f(f(y))$$and by switching $x$ and $y$ in the statement above we get that : $\quad \forall x,y \in \mathbb{R}^{+}$ : $$f(x)+3x+f(y)f(f(y))=f(y)+3y+f(x)f(f(x))$$$$\implies f(x)+ 3x-f(x)f(f(x))= f(y)+ 3y-f(y)f(f(y))$$ Which implies that $f(x)f(f(x))-f(x)-3x$ is constant for all positive real $x$ but since $f(x)\ge x$ we get that: $f(x)f(f(x))-f(x)-3x\ge f(x)^2-f(x)-3x\ge f(x)^2-4f(x)$. The latter is a quadratic polynomial and in the problem we know that by putting $x$ a very big number, $f(f(x)+y)$ also gets very big so the quadratic polynomial:$f(x)^2-4f(x)$ can get very large $\implies f(x)f(f(x))-f(x)-3x$ can't be constant$\implies$ there is no function satisfying the equation. $Q.E.D$

Part of it a it hard to read, but you can click on the mashed-up words to see what is written.

GrizzlyTurtle wrote: Part of it a it hard to read, but you can click on the mashed-up words to see what is written. thanks for the notice is it better now?

Yes, much.

it is very doable problem and it is very easy $P(1,1)$ $f(1+f(1))=f(1)+3+f(1)$ it motivates us to look some inequalities $f(1)<1$ gives contradicition so let's take it to another level Assume that $f(x)<x$ for all $x$ $P(x,x-f(x)$ gives contradicition Thus we have $f(x) \geq x$ we can change LHS $f$ like take some constant $c$ which is sufficiently small but not $0$ $P(x,c)$ and $P(c,f(x)$ gives us $x$ has border over some constant $c$ if we take $x$ very very big it gives us contradicition. and we are done

if there exist $x$ such that $x> f(x)$, we can pick $y=x-f(x)$. This mean that $3x+(x-f(x))f(x-f(x))=0$, so it contradicts ($x>0$). Therefore, $f(x) \geq x$ for all $x$. After this, we can pick $P(x;f(y))$,so $f(f(x)+f(y))=f(x)+3x+f(y)f(f(y))=f(y)+3y+f(x)f(f(x))$ (After switch $x$ and $y$) $f(x)f(f(x))-f(x)-3x=f(y)f(f(y))-f(y)-3y=f(y)(f(f(y))-1)-3y>y(f(f(y))-4)>y(y-4)$ We can pick very big $y$ for every $x$,such that $y(y-4)>LHS$. It contradicts

OmarGadimli wrote: if there exist $x$ such that $x> f(x)$, we can pick $y=x-f(x)$. This mean that $3x+(x-f(x))f(x-f(x))=0$, so it contradicts ($x>0$). Therefore, $f(x) \geq x$ for all $x$. After this, we can pick $P(x;f(y))$,so $f(f(x)+f(y))=f(x)+3x+f(y)f(f(y))=f(y)+3y+f(x)f(f(x))$ (After switch $x$ and $y$) $f(x)f(f(x))-f(x)-3x=f(y)f(f(y))-f(y)-3y=f(y)(f(f(y))-1)-3y>y(f(f(y))-4)>y(y-4)$ We can pick very big $y$ for every $x$,such that $y(y-4)>LHS$. It contradicts İ think there is a gap in your inequality, the one before the last one. it is not true for the case $f(f(y))<1$.

Tellocan wrote: OmarGadimli wrote: if there exist $x$ such that $x> f(x)$, we can pick $y=x-f(x)$. This mean that $3x+(x-f(x))f(x-f(x))=0$, so it contradicts ($x>0$). Therefore, $f(x) \geq x$ for all $x$. After this, we can pick $P(x;f(y))$,so $f(f(x)+f(y))=f(x)+3x+f(y)f(f(y))=f(y)+3y+f(x)f(f(x))$ (After switch $x$ and $y$) $f(x)f(f(x))-f(x)-3x=f(y)f(f(y))-f(y)-3y=f(y)(f(f(y))-1)-3y>y(f(f(y))-4)>y(y-4)$ We can pick very big $y$ for every $x$,such that $y(y-4)>LHS$. It contradicts İ think there is a gap in your inequality, the one before the last one. it is not true for the case $f(f(y))<1$. $f(f(y))$ > $f(y)$ > $y$ so if $y>1$ it means that $f(f(y))$ > 1