SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Replacing $x=y$ in i) and ii), we get $f(2x)=4f(x)$, $\forall x\in {\mathbb{R}}^{+}$. Therefore, by induction, we can prove that $$f\left(2^{n}x\right)=4^{n}f(x),\forall n\in {\mathbb{Z}}^{+},x\in {\mathbb{R}}^{+}.$$ Now, by substituting $y=2x$ in i) and ii), we get $10f(x)\le f(3x)\le 9f(x)$, $\forall x\in {\mathbb{R}}^{+}$. It follows that $f(x)\le 0,\forall x\in {\mathbb{R}}^{+}$. Therefore, from ii), we have $$xyf(x+y)\le (x+y)yf(x)\le xyf(x),$$ or $$f(x+y)\le f(x),\forall x,y\in {\mathbb{R}}^{+}.$$ Next, we rewrite the condition ii) as $$\frac{f(x+y)}{x+y}\le \frac{f(x)}{x}+\frac{f(y)}{y}.$$ By substituting $y=2x,3x,\dots ,nx\left(n\in {\mathbb{Z}}^{+}\right)$, we can prove by induction that $$f(nx)\le n^{2}f(x),\forall n\in {\mathbb{Z}}^{+},x\in {\mathbb{R}}^{+}.\text{\hspace{1em}(*)}$$ From these inequalities, we get $$\begin{array}{rl}\frac{2^{n}f(x)}{x}& =\frac{f\left(2^{n}x\right)}{2^{n}x}=\frac{f\left(\left(2^n-n\right)x+nx\right)}{\left(2^n-n\right)x+nx}\\ & \le \frac{f\left(\left(2^n-n\right)x\right)}{\left(2^n-n\right)x}+\frac{f(nx)}{nx}\\ & \le \frac{\left(2^n-n\right)f(x)}{x}+\frac{nf(x)}{x}=\frac{2^{n}f(x)}{x}\end{array}$$ for any positive integer $n$ and any positive real number $x$. Therefore, the equality cases in (*) must hold, i.e. $$f(nx)=n^{2}f(x),\forall n\in {\mathbb{Z}}^{+},x\in {\mathbb{R}}^{+}.$$ From this, we can easily prove that $f(x)=k{x}^2,\forall x\in {\mathbb{Q}}^{+}$ (where $k=f(1)$, and $k\le 0$). And then, by combining this result with $f(x+y)\le f(x),\forall x,y>0$, we conclude that $f(x)=k{x}^2,\forall x\in {\mathbb{R}}^{+}$. It is easy to check that this function satisfy the given conditions.