pablock wrote:

Let $\mathbb{R}_{>0}$ be the set of the positive real numbers. Find all functions $f:\mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that $f(xy+f(x))=f(f(x)f(y))+x$ for all positive real numbers $x$ and $y$ .

Solution. We prove three claims.

Claim 1. $f(x)\geq x$ .
Proof. Assume there exists $a$ so that $f(a)<a$ . Then, $P\left(x,\frac{x-f(x)}{x}\right)$ gives $f(M)=f(x)-x>0$ , a contradiction.

Claim 2. $f(x)<1$ for all $x<1$ .
Proof. For $x<1$ , choose $y$ so that $xy+f(x)=y$ . Then, $f(y)=f(xy+f(x))=f(f(x)f(y))+x>f(x)f(y)+x>f(x)f(y)$ and so we are done.

Claim 3. $\lim_{t\rightarrow0}f(t)=0$ .
Proof. Choose $0<\epsilon<1$ arbitrarily small. Then $1-\epsilon\in(0,1)$ . Also choose $\lambda$ such that $(1-\epsilon)\lambda+f(1-\epsilon)<1$ Then, $x=1-\epsilon$ and $y=\lambda$ gives $1>f(x\lambda+f(x))=f(f(x)f(\lambda))+x=f(f(x)f(\lambda))+1-\epsilon$ so $0<f(f(x)f(\lambda))<\epsilon$ and since $\epsilon$ is arbitrarily small, we have that $f(f(x)f(\lambda))$ tends to $0$ . Now, $P(x,\lambda)$ gives $x\leq f(x)<x\lambda+f(x)\leq f(x\lambda+f(x))=f((x)f(\lambda))+x$ and since $f(f(x)f(\lambda))$ tends to $0$ , $f(x)=x$ . Switching $x$ and $\lambda$ gives $f(\lambda)=\lambda$ . Therefore, we have that $f(x)f(\lambda)=x\lambda$ (since $f(x)=x$ for $x<1$ ) which tends to $0$ , if we choose $\lambda$ sufficiently small. Therefore, as $f(x)f(\lambda)$ tends to $0$ , it's image also tends to $0$ , as desired.

Finally, if we fix $x=x_0$ and choose $y=\delta$ sufficiently small, then $x\leq f(x)<xy+f(x)\leq f(xy+f(x))=x+f(f(x)f(y))$ Also we have $\lim_{y\rightarrow0}f(y)=0$ so $\lim_{y\rightarrow0}f(f(x)f(y))=0$ and so $f(x)=x$ (since the first inequality is actually an equality)