Cesko-Slovacko-Poljsko 2006

Let the numbers be $p$, $q$, $r$, $s$. Up to a permutation we may assume that $p\ge q\ge r\ge s$. We first consider the case where $p+q\ge 5$. Then $$p^2+q^2+2pq\ge 25=4+(p^2+q^2+r^2+s^2)\ge 4+p^2+q^2+2rs,$$ which is equivalent to $pq-rs\ge 2$.

Assume now that $p+q<5$; then $$4<r+s\le p+q<5.\text{\hspace{1em}(1)}$$ Observe that $$(pq+rs)+(pr+qs)+(ps+qr)=\frac{(p+q+r+s{)}^2-(p^2+q^2+r^2+s^2)}{2}=30.$$ Moreover, $$pq+rs\ge pr+qs\ge ps+qr,$$ because $(p-s)(q-r)\ge 0$ and $(p-q)(r-s)\ge 0$. We conclude that $pq+rs\ge 10$. From (1), we get $0\le (p+q)-(r+s)<1$, therefore $$(p+q{)}^2-2(p+q)(r+s)+(r+s{)}^2<1.$$ Adding this to $(p+q{)}^2+2(p+q)(r+s)+(r+s{)}^2=9^2$ gives $$(p+q{)}^2+(r+s{)}^2<41.$$ Therefore $$41=21+2\cdot 10\le (p^2+q^2+r^2+s^2)+2(pq+rs)=(p+q{)}^2+(r+s{)}^2<41,$$ which is a contradiction.

Second solution.

From $a+b+c+d=9$ with ordering $a\ge b\ge c\ge d$ we have $$\frac{a+b}{2}=\frac{9}{4}+{\epsilon }_1,\frac{c+d}{2}=\frac{9}{4}-{\epsilon }_1$$ for some ${\epsilon }_1\ge 0$. Thus $$a=\frac{9}{4}+{\epsilon }_1+{\epsilon }_2,b=\frac{9}{4}+{\epsilon }_1-{\epsilon }_2,c=\frac{9}{4}-{\epsilon }_1+{\epsilon }_3,d=\frac{9}{4}-{\epsilon }_1-{\epsilon }_3$$ for some ${\epsilon }_2,{\epsilon }_3\ge 0$. From $b\ge c$ we get $${\epsilon }_1-{\epsilon }_2\ge -{\epsilon }_1+{\epsilon }_3\text{or}{\epsilon }_2+{\epsilon }_3\le 2{\epsilon }_1.$$ From $$\begin{gathered} 21=(a^2+b^2)+(c^2+d^2)=2\cdot {\left(\frac{9}{4}+{\epsilon }_1\right)}^2+2{\epsilon }_2^2+2\cdot {\left(\frac{9}{4}-{\epsilon }_1\right)}^2+2{\epsilon }_3^2= \\ =4\cdot {\left(\frac{9}{4}\right)}^2+4{\epsilon }_1^2+2{\epsilon }_2^2+2{\epsilon }_3^2=20+\frac{1}{4}+2\cdot (2{\epsilon }_1^2+{\epsilon }_2^2+{\epsilon }_3^2) \end{gathered}$$ we conclude that non-negative numbers ${\epsilon }_i$ satisfy $$2{\epsilon }_1^2+{\epsilon }_2^2+{\epsilon }_3^2=\frac{3}{8}.\text{\hspace{1em}(2)}$$ Using ${\epsilon }_2+{\epsilon }_3\le 2{\epsilon }_1$ and ${\epsilon }_2^2+{\epsilon }_3^2\le ({\epsilon }_2+{\epsilon }_3{)}^2$ from (2) we obtain $$\frac{3}{8}\le 2{\epsilon }_1^2+({\epsilon }_2+{\epsilon }_3{)}^2\le 2{\epsilon }_1^2+4{\epsilon }_1^2=6{\epsilon }_1^2.$$ Hence ${\epsilon }_1^2\ge (1/6)\cdot (3/8)=1/16$ or ${\epsilon }_1\ge 1/4$. For $ab-cd$ we have $$ab-cd={\left(\frac{9}{4}+{\epsilon }_1\right)}^2-{\epsilon }_2^2-{\left(\frac{9}{4}-{\epsilon }_1\right)}^2+{\epsilon }_3^2=9{\epsilon }_1-{\epsilon }_2^2+{\epsilon }_3^2.$$ Substituting for ${\epsilon }_2^2$ from (2) we get $$ab-cd=9{\epsilon }_1-\left(\frac{3}{8}-2{\epsilon }_1^2-{\epsilon }_3^2\right)+{\epsilon }_3^2=9{\epsilon }_1+2{\epsilon }_1^2-\frac{3}{8}+2{\epsilon }_3^2\ge 9\cdot \frac{1}{4}+2\cdot \frac{1}{16}-\frac{3}{8}=2.$$ This ends the proof.