Hopeooooo wrote: Find all functions $f:R->R$ satisfying that for every $x$ (real number): $f(x)(1+|f(x)|)\geq x \geq f(x(1+|x|))$ Let $t\le 0$ Setting $x=\frac{1-\sqrt{1-4t}}2\le 0$ in $x\ge f(x(1+|x|)$, we get $f(t)\le \frac{1-\sqrt{1-4t}}2$ $\forall t\le 0$ This implies $f(t)\le 0$ and $f(t)(1+|f(t)|)\ge t$ becomes $f(t)^2-f(t)+t\le 0$ And so $\frac {1-\sqrt{1-4t}}2\le f(t)\le 0$ And so $f(x)=\frac{1-\sqrt{1-4x}}2$ $\forall x\le 0$ Let $t>0$ Setting $x=\frac{-1+\sqrt{1+4t}}2\ge 0$ in $x\ge f(x(1+|x|)$, we get $f(t)\le \frac{-1+\sqrt{1+4t}}2$ $\forall t\ge 0$ If $f(t)\le 0$ then $f(t)(1+|f(t)|)\ge t$ would imply $t\le 0$, impossible. So $f(t)>0$ And we have $0< f(t)\le \frac{-1+\sqrt{1+4t}}2$ Then $f(t)(1+|f(t)|)\ge t$ becomes $f(t)^2+f(t)-t\ge 0$ and so $f(t)\ge \frac{-1+\sqrt{1+4t}}2$ And so $f(x)=\frac{-1+\sqrt{1+4x}}2$ $\forall x> 0$ And so $\boxed{f(0)=0\text{ and }f(x)=\frac{|x|}{2x}\left(\sqrt{1+4|x|}-1\right)\quad\forall x\ne 0}$

Let $-a = x \le 0$ and $k = x(1+|x|) = -a - a^2$ so $a^2 + a + k = 0 \implies x=\frac{1\pm \sqrt{1-4k}}2$. Case $1$ : $ k \le 0$. $x : \frac{1-\sqrt{1-4k}}2 \implies f(k)\le \frac{1-\sqrt{1-4k}}2 \implies f(k) \le 0 \implies f(k)^2 - f(k) + k \le 0 \implies \frac {1-\sqrt{1-4k}}2\le f(k) \implies f(k) = \frac{1-\sqrt{1-4k}}2 \forall k\le 0 \implies f(x) = \frac{1-\sqrt{1-4x}}2 \forall x\le 0$. Case $2$ : $k > 0$. $x : \frac{-1+\sqrt{1+4k}}2 \implies f(k)\le \frac{-1+\sqrt{1+4k}}2$. $f(x)(1+|f(x)|) \ge x \implies f(k) > 0 \implies f(k)^2 + f(k) - k \ge 0 \implies f(k)\ge \frac{-1+\sqrt{1+4k}}2 \implies f(k)=\frac{-1+\sqrt{1+4k}}2 \forall k> 0 \implies f(x)=\frac{-1+\sqrt{1+4x}}2 \forall x> 0$ so now we have $f$ for $x > 0,x = 0, x < 0$. we're Done.