SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Let $x=y$ in (2), we have $f\left(x^2\right)\ge f^2(x)\ge 0$. This implies that $f(x)\ge 0$ for all $x\ge 0$.

Let $x=0$ in (1), we have $f(1)\ge f(0)+1\ge 1$.

Let $x=y=1$ in (1), we have $f(1)\ge f^2(1)$, i.e., $0\le f(1)\le 1$. These inequalities imply that $f(1)=1$, and hence $f(0)=0$.

From (1) and (2), by induction, we have $$f(x+n)\ge f(x)+n\text{\hspace{1em}(*)}$$ for all $x\in \mathbb{R}$, $n\in \mathbb{N}$, and $$f\left(x^n\right)\ge f^n(x)\text{\hspace{1em}(**)}$$ for all $x\ge 0$ and $n\in {\mathbb{N}}^{\ast }$.

Using $(\ast )$, for each $x>0$, we have $$f(x)=f(\{x\}+\lfloor x\rfloor )\ge f(\{x\})+\lfloor x\rfloor \ge \lfloor x\rfloor >x-1.$$ From $(\ast \ast )$ and (2), for each $x>1$ and $n\in {\mathbb{N}}^{\ast }$, we have $$\left(x^n-1\right)f^n\left(\frac{1}{x}\right)\le f\left(x^n\right)f\left(\frac{1}{x^n}\right)\le f(1)=1.$$ Hence, $$f\left(\frac{1}{x}\right)\le \frac{1}{\sqrt[n]{x^n-1}}$$ for all $x>1$, $n\in {\mathbb{N}}^{\ast }$. Fix $x>1$ and let $n\to +\infty$, we obtain $f\left(\frac{1}{x}\right)\le \frac{1}{x}$, or (note that $f(0)=0$ and $f(1)=1$), $$f(x)\le x$$ for $x\in [0,1]$.

On the other hand, we can prove by induction from (1) that $$f(x-n)\le f(x)-n$$ for $n\in \mathbb{N}$. For each $x<0$, we have $\lfloor x\rfloor <0$ so $$f(x)=f(\{x\}+\lfloor x\rfloor )\le f(\{x\})+\lfloor x\rfloor \le \{x\}+\lfloor x\rfloor =x.$$ In (2), set $x=y=t$ with $t\in [-1,0]$ then $$f^2(t)\le f\left(t^2\right)\le t^2$$ This implies that $t\le f(t)\le -t$. Hence, $f(t)=t$ for all $t\in [-1,0]$.

Now, for each $x<-1$, we have $$\frac{1}{x^2}{f}^2(x)=f^2(x)f^2\left(\frac{1}{x}\right)\le f\left(x^2\right)f\left(\frac{1}{x^2}\right)\le f(1)=1$$ so $f^2(x)\le x^2$, implying that $x\le f(x)\le -x$. So $f(x)=x$ for all $x\le 0$.

Let $y=-1$ in (2) and consider $x>0$, we have $$-x=f(-x)\ge f(x)f(-1)=-f(x)$$ implying $f(x)\ge x$ for all $x>0$. Hence, we have $f(x)=x$ for all $x\in (0,1)$.

Finally, suppose that $x>1$, we have $$\frac{1}{x}f(x)=f(x)f\left(\frac{1}{x}\right)\le f(1)=1$$ so $f(x)\le x$. Since $f(x)\ge x$, we have $f(x)=x$ for all $x>1$.

Inclusion, $f(x)=x$ for every $x$ is the only function satisfying given conditions.