Czech-Polish-Slovak Match

Let us set $$f(x,y)=(x+y)(x^2+y^2).$$ We'll show that for any real $x$ the inequality $y\ge z$ implies $f(x,y)\ge f(x,z)$. After subtraction we see that $$f(x,y)-f(x,z)=\frac{1}{2}(y-z)((x+y{)}^2+(y+z{)}^2+(z+x{)}^2)\ge 0.$$ Moreover, equality occurs when $y=z$ or $x=y=z=0$, so either way it implies $y=z$.

We can rewrite the system (implicitly using the symmetry of $f$) to the form: $$\begin{array}{rl}f(a,b)& =f(c,d)\\ f(a,c)& =f(b,d)\\ f(a,d)& =f(b,c)\end{array}$$ Now we can see that the system is symmetric in variables $a,b,c,d$ and may assume $a=\max \{a,b,c,d\}$. We then write the chain of (in)equalities $$f(c,d)=f(a,b)\ge f(c,b)=f(a,d)\ge f(b,d)=f(a,c)\ge f(d,c)$$ and since we in fact have equality everywhere, we deduce $a=b=c=d$. All such quadruplets clearly satisfy the system so the problem is solved. □