Ignore this

TalTheMagician wrote: Someone correct me if I have any mistakes. The fact that: $$f(0) \leqslant -f(x)^2-f(x)$$implies $f(x)^2+f(x) \leqslant -f(0)$. Your next line $x(f(x)+1) \leqslant x(f(0)+1)+1$ would hold only if $x \geqslant 0$.

Let the given assertion be P(x,y) P(0,y) gives f(y)^2+f(y)+f(0)<=0 and we deduce that f is bounded. On the other hand pick some y with f(y)!=-1 and vary x to get that f is not bounded from below.

Mikeglicker wrote: Let the given assertion be P(x,y) P(0,y) gives f(y)^2+f(y)+f(0)<=0 and we deduce that f is bounded. On the other hand pick some y with f(y)!=-1 and vary x to get that f is not bounded from below. Well, you can't pick f(y)=-1 because there might not be a $y$ that $f(y)=-1$

For the sake of contradiction, suppose that there exists a non-constant function $f: \mathbb{R} \to \mathbb{R}$ such that \[ f(x + f(y)) \leq x f(y) + x \quad \forall x, y \in \mathbb{R} \]Putting $x = -f(y)$ we can see that $f(y)^2 + f(y) + f(0) \leq 0 \, \forall y \in \mathbb{R}$. Since $f$ is non-constant, there exists $y_0 \in \mathbb{R}$ such that $f(y_0) \neq -\frac{1}{2}$, so if $f(0) \geq \frac{1}{4}$ then \[ f(y_0)^2 + f(y_0) + f(0) \geq \left( f(y_0) + \frac{1}{2} \right)^2 > 0 \]a contradiction. So $f(0) < \frac{1}{4}$ and we can deduce that $\ell \leq f(y) \leq u \, \forall y \in \mathbb{R}$ where \[ \ell = \frac{-1 - \sqrt{1 - 4f(0)}}{2} < -1 \text{ and } u = \frac{-1 + \sqrt{1 - 4f(0)}}{2} \]For $x > 0$ we have $f(y) + 1 \geq \frac{f(x + f(y))}{x} \geq \frac{\ell}{x}$. Letting $x \to +\infty$ we obtain that $f(y) \geq -1 \, \forall y \in \mathbb{R}$. Using again the hypothesis that $f$ is non-constant, we can choose $y^*$ such that $f(y^*) \neq -1$, which, combining with what we have just proved, implies that $f(y^*) > - 1$. Now letting $y = y^*$ we have \[ -1 \leq f(x + f(y^*) ) \leq x(f(y^*) + 1) \quad \forall x \in \mathbb{R} \]which yields a contradiction when $x \to -\infty$.

a22886 wrote: Mikeglicker wrote: Let the given assertion be P(x,y) P(0,y) gives f(y)^2+f(y)+f(0)<=0 and we deduce that f is bounded. On the other hand pick some y with f(y)!=-1 and vary x to get that f is not bounded from below. Well, you can't pick f(y)=-1 because there might not be a $y$ that $f(y)=-1$ I think they mean $f(y) \not= -1$, and this is possible because $f$ is not constant. The solution seems essentially correct, except for a typo: It should be $P(-f(y) , y)$ instead of $P(0,y)$.

$P(-f(y) , y)$ $f(y)^2 + f(y) + f(0) \leq 0 \, \forall y \in \mathbb{R}$ Exist $a$ and $b$ $$a\leq f(x)\leq b$$$$x(f(y)+1)\ge a$$$y$-fixed. $$xd\ge a$$Contradiction