Do you have the whole test?

DanDumitrescu wrote: Determine functions $f : \mathbb{R} \rightarrow \mathbb{R},$ with property that \[ f(f(x)) + y \cdot f(x) \le x + x \cdot f(f(y)), \] for every $x$ and $y$ are real numbers. Let $P(x,y)$ be the assertion $f(f(x))+yf(x)\le x+xf(f(y))$ $P(0,x)$ $\implies$ $f(f(0))+xf(0)\le 0$ $\forall x$ and so $f(0)=0$ $P(x,0)$ $\implies$ $f(f(x))\le x$ $P(1,x)$ $\implies$ $f(f(1))+xf(1)\le 1+f(f(x))$ $\le 1+x$ and so $f(1)=1$ And previous line becomes $x\le f(f(x))$ and so $f(f(x))=x$ Then $P(x,y)$ becomes $yf(x)\le xy$ $\forall x,y$ and so $\boxed{f(x)=x\quad\forall x}$, which indeed fits.

pco wrote: DanDumitrescu wrote: Determine functions $f : \mathbb{R} \rightarrow \mathbb{R},$ with property that \[ f(f(x)) + y \cdot f(x) \le x + x \cdot f(f(y)), \] for every $x$ and $y$ are real numbers. Let $P(x,y)$ be the assertion $f(f(x))+yf(x)\le x+xf(f(y))$ $P(0,x)$ $\implies$ $f(f(0))+xf(0)\le 0$ $\forall x$ and so $f(0)=0$ $P(x,0)$ $\implies$ $f(f(x))\le x$ $P(1,x)$ $\implies$ $f(f(1))+xf(1)\le 1+f(f(x))$ $\le 1+x$ and so $f(1)=1$ And previous line becomes $x\le f(f(x))$ and so $f(f(x))=x$ Then $P(x,y)$ becomes $yf(x)\le xy$ $\forall x,y$ and so $\boxed{f(x)=x\quad\forall x}$, which indeed fits. Why f(1) = 1 ? Sorry if it was obvious

demmy wrote: Why f(1) = 1 ? Sorry if it was obvious We got $x(f(1)-1)\le 1-f(f(1))$ $\forall x$ If $f(1)\ne 1$, choosing $x\to\infty$ with same sign as $f(1)-1$ set $LHS\to+\infty$ and so inequality wrong

pco wrote: demmy wrote: Why f(1) = 1 ? Sorry if it was obvious We got $x(f(1)-1)\le 1-f(f(1))$ $\forall x$ If $f(1)\ne 1$, choosing $x\to\infty$ with same sign as $f(1)-1$ set $LHS\to+\infty$ and so inequality wrong Thank you.