26th Turkish Mathematical Olympiad

Answer: $(x,y)=(-1/3,-1/3)$, $(5/3,5/3)$. Firstly, we obtain that $$x^2+y^2+x+y-xy(x+y)+2=-(1-x)(1-y)(x+y+2)$$ and hence we get $$(1-x)(1-y)(x+y+2)=\frac{64}{27}.$$ Let $k^3=1-x$, ${\ell }^3=1-y$, $m^3=x+y+2$. Then we have $$k^3+{\ell }^3+m^3=4\text{and}k\ell m=\frac{4}{3}.$$ This shows that $k^3+{\ell }^3+m^3=3k\ell m$ and it can be expressed as $$(k+\ell +m)((k-\ell {)}^2+(\ell -m{)}^2+(m-k{)}^2)=0.$$ The last equality implies that either $k+\ell +m=0$ or $k=\ell =m$. For $k=\ell =m$, we obtain the solution $x=y=-1/3$. Let $k+\ell +m=0$. In this case using $k\ell m=4/3$, we get $k\ell (k+\ell )=-4/3$. Consider the condition $|xy|=|(1-k^3)(1-{\ell }^3)|\le 25/9$. Define $u=k+\ell$, $v=k\ell$. In this case, we get $$uv=-\frac{4}{3}\text{and}|(1-k^3)(1-{\ell }^3)|=|u^3-v^3+3|\le \frac{25}{9}.$$ These two conditions imply that $$-\frac{52}{9}\le u^3-v^3=u^3+\frac{64}{27u^3}\le -\frac{2}{9}(\ast )$$ By AM-GM, we get $u^2\ge 4v=-16/(3u)$ which is equivalent to $u>0$ or $u^3\le -16/3$. From (), we conclude that $u<0$ and hence $u^3\le -16/3$. In this case since $u^3+16/3\le 0$ and $4/(9u^3)+1\ge 11/12>0$ we get that $$u^3+\frac{64}{27u^3}+\frac{52}{9}=\left(u^3+\frac{16}{3}\right)\left(\frac{4}{9u^3}+1\right)\le 0$$ Therefore, () holds only if $u^3=-16/3$ and $u^2=4v$ which implies that $k=\ell =-\sqrt[3]{2}/3$. This yields the solution $x=y=5/3$. Both obtained solutions satisfy the problem conditions.

---

Alternative solution.

Let $x+y=a$, $xy=b$. Then by AM-GM, we have $a^2\ge 4b$ and hence $$x^2+y^2+x+y-xy(x+y)=a^2+a-b(a+2)=-\frac{10}{27},$$ $$b=\frac{a^2+a+\frac{10}{27}}{a+2}\le \frac{a^2}{4},$$ $$|b|=\left|\frac{a^2+a+\frac{10}{27}}{a+2}\right|\le \frac{25}{9}.$$ Therefore, we get $$\frac{a^2}{4}-\frac{a^2+a+\frac{10}{27}}{a+2}=\frac{{\left(a+\frac{2}{3}\right)}^2\left(a-\frac{10}{3}\right)}{4(a+2)}\ge 0(1),$$ $$\frac{a^2+a+\frac{10}{27}}{a+2}-\frac{25}{9}=\frac{\left(a-\frac{10}{3}\right)\left(a+\frac{14}{9}\right)}{a+2}\le 0(2),$$ $$\frac{a^2+a+\frac{10}{27}}{a+2}+\frac{25}{9}=\frac{{\left(a+\frac{17}{9}\right)}^2+\frac{191}{81}}{a+2}\ge 0(3).$$ Using (1), (2) and (3), we obtain the following solutions: $$(1)⟺a<-2\text{or}a\ge \frac{10}{3}\text{or}a=-\frac{2}{3},$$ $$(2)⟺a<-2\text{or}-\frac{14}{9}\le a\le \frac{10}{3},$$ $$(3)⟺a>-2.$$ Hence, the only solutions are $a=-\frac{2}{3}$ and $a=\frac{10}{3}$. In both cases, $a^2=4b$ and it follows that $x=y=\frac{a}{2}$. The obtained solutions $(x,y)=(-1/3,-1/3)$, $(5/3,5/3)$ satisfy the problem conditions.