Existence of two real numbers satisfying $f(x-f(y))>yf(x)+x$

Let $f:\mathbb{R} \longrightarrow \mathbb{R}$ be a function. Is it always the case that for some $x,y \in \mathbb R$, the inequality $f(x-f(y))>yf(x)+x$ holds?

Thanks in advance.

Solution 1:

Suppose $f(x− f(y)) \leqslant yf(x)+x$ for all $x,y \in \mathbb{R}$, and write $f(0)=\alpha,\;f(-1)=\beta$.

$\begin{aligned}\textbf{Claim}:&\quad \text{(a)}:\quad f(x)\leqslant x+\alpha\\ &\;\;\,\text{ (b)}:\quad x>0\Rightarrow f(f(x))+1\geqslant 0\end{aligned}$

$\textit{Proof}:$ Let $y = 0$ and $x\mapsto x+\alpha$ and we immediately have claim $\text{(a)}$. Next, set $x= f(y)$ then $0\leqslant y f ( f (y)) + y $ by $\text{(a)}$, whence claim $\text{(b)}$.

$\begin{aligned}\textbf{Claim}:\quad\text{(c)}:\quad f(x)\leqslant 0\end{aligned}$

$\textit{Proof}:$ For the sake of contradiction suppose $f(\xi)>0$ for some $\xi$. For all $x<\xi−\alpha$ we have $\;f(x)−\alpha\leqslant x< \xi−\alpha$ by $\text{(a)}$. In other words, $0<\xi-f(x)$. Utilising $\text{(a)}$ and $\text{(b)}:$$$\begin{aligned}0 \underset{\text{(b)}}\leqslant f\big(f\big(\xi−f(x)\big)\big)+1\underset{\text{(a)}}\leqslant f\big(\xi− f(x)\big)+ \alpha+1 \leqslant xf(\xi)+\alpha+ \xi+1\end{aligned}$$ Equivalently, $-xf(\xi) \leqslant (1+\xi+\alpha)$ which is absurd for $-x$ sufficiently large, whence $\text{(c)}$. This is enough material for a contradiction: $\text{(a)}$ and $\text{(c)}$ yield $ f(x)\leqslant x\;\text{(d)}$. For $x>0:$ $$\begin{aligned} \beta= f\Big((f(x)−1)−f(x)\Big) &\leqslant xf\big(f(x)−1\big)+ f(x)−1 \\&\underset{\text{(d)}}\leqslant (x+1)\big(f(x)−1\big)\underset{\text{(c)}}\leqslant −(x+1)\end{aligned}\quad\Big(\text{i.e.}\quad x\leqslant -(\beta+1)\Big)$$ This is nonsense when $x$ is sufficiently large $-$ your conjecture is correct.