Middle European Mathematical Olympiad T-2

Problem: Determine all functions $f:\mathbb{R}\to \mathbb{R}$ such that: $xf(xy)+xyf(x)\ge f(x^2)f(y)+x^2y$ holds for all $x,y\in \mathbb{R}$.
Solution: Let $(x,y)$ denote the assertion $xf(xy)+xyf(x)\ge f(x^2)f(y)+x^2y$.
$(x,x)\implies (x^2-f(x^2))(x-f(x))\le 0,$denote this by $(1)$
$(0,0)\implies f(0)=0$
$(1,1)\implies f(1)=1$.
$(x,1)\implies 2xf(x)\ge f(x^2)+x^2\implies f(x)^2-f(x^2)\ge (f(x)-x)^2,$denoted this by $(2)$.
Now, suppose that there exists $a\ge 0$ such that $a>f(a)$.
$(1)\implies (a^2-f(a^2))(a-f(a))\le 0\implies a^2-f(a^2)\le 0\implies -f(a^2)\le -a^2$.
$(2)\implies (f(a)-a)^2\le f(a)^2-f(a^2)\le f(a)^2-a^2=(f(a)-a)(f(a)+a)$
As $f(a)-a<0$, we get $f(a)-a\ge f(a)+a\implies 2a\le 0$, which isn't true. So $f(x)\ge x$ for every $x\ge 0$.
That means $f(x^2)\ge x^2\forall x$.
$(1)\implies x\ge f(x)\forall x\implies \boxed{f(x)=x\forall x\ge 0}$
$(x<0,y<0)\implies x(yf(x)-xf(y))\ge 0\implies yf(x)\le xf(y)$, and let that be $(3)$.
$(y,x)\rightarrow (3)\implies xf(y)\le yf(x)\implies xf(y)=yf(x)\implies f(x)=cx\forall x<0$.
$(2)\implies 2xf(x)\ge f(x^2)+x^2\ge 2x^2\implies x(f(x)-x)\ge 0\implies f(x)\le x\forall x<0$
So, $\boxed{f(x)=cx\forall x<0}$ with $c\ge 1$.
It's easy to see that:
$\boxed{f(x)=\left\{\begin{array}{I} x\text{ if }x\ge 0\\cx\text{ if }x<0 \end{array}\right.,c\ge 1}$
Source: https://artofproblemsolving.com/community/c6h607043_functional_inequality