Vietnam Mathematical Olympiad

Write the given condition in the form $$x+y=3(\sqrt{x+1}+\sqrt{y+2}).$$ Denote by $G$ the set of values of $P$. It is easily seen that: $a\in G\iff a$ is a real number so that the following system of equations (with unknowns $x,y$) has solutions $$\{\begin{array}{l}3(\sqrt{x+1}+\sqrt{y+2})=a\\ x+y=a.\end{array}(I)$$ By putting $u=\sqrt{x+1}$ and $v=\sqrt{y+2}$, from the system (I), we get the following system of equations (with unknowns $u,v$) $$\{\begin{array}{l}3(u+v)=a\\ u^2+v^2=a+3.\end{array}(II)$$ which is equivalent to $$\{\begin{array}{l}u+v=\frac{a}{3}\\ uv=\frac{1}{2}\left(\frac{a^2}{9}-a-3\right).\end{array}$$ Therefore, The system (I) has solutions $\iff$ the system (II) has the solutions $(u,v)$ with $u,v$ non negative $\iff$ the equation (with unknown $t$) $18t^2-6at+a^2-9a-27=0$ has two non negative roots $\iff$ $$\{\begin{array}{l}-a^2+18a+54\ge 0\\ a\ge 0\\ a^2-9a-27\ge 0\end{array}\iff \frac{9+3\sqrt{21}}{2}\le a\le 9+3\sqrt{15}.$$ Hence $G=\left[\frac{9+3\sqrt{21}}{2};9+3\sqrt{15}\right]$ and so $\min P=\frac{9+3\sqrt{21}}{2}$, $\max P=9+3\sqrt{15}$.