SELECTION EXAMINATION

Since $x,y,z$ are greater than $3$, it follows that $y+z-2,z+x-4,x+y-6$ are positive. Thus, from Cauchy-Schwarz inequality we get: $$\begin{gathered} \left(\frac{(x+2{)}^2}{y+z-2}+\frac{(y+4{)}^2}{x+z-4}+\frac{(z+6{)}^2}{x+y-6}\right)\left((y+z-2)+(x+z-4)+(x+y-6)\right)\ge (x+y+z+12{)}^2 \\ \iff \frac{(x+2{)}^2}{y+z-2}+\frac{(y+4{)}^2}{x+z-4}+\frac{(z+6{)}^2}{x+y-6}\ge \frac{1}{2}\cdot \frac{(x+y+z+12{)}^2}{(x+y+z-6)}. \end{gathered}$$ From the hypothesis of the problem it follows that $$\frac{(x+y+z+12{)}^2}{(x+y+z-6)}\le 72,(1)$$ where as the equality holds when: $$\frac{x+2}{y+z-2}=\frac{y+4}{x+z-4}=\frac{z+6}{x+y-6}=\lambda \iff \{\begin{array}{l}\lambda (y+z)-x=2(\lambda +1)\\ \lambda (x+z)-y=4(\lambda +1)\\ \lambda (x+y)-z=6(\lambda +1)\end{array}.(2)$$ ---

Moreover, we observe that: $$\frac{(x+y+z+12{)}^2}{(x+y+z-6)}=\frac{(x+y+z+12{)}^2}{(x+y+z+12)-18}=\frac{{\omega }^2}{\omega -18},$$ where we have put $\omega =x+y+z+12$. Since we have $$\frac{{\omega }^2}{\omega -18}\ge 4\cdot 18=72\iff {\omega }^2-4\cdot 18\omega +4\cdot 18^2\ge 0\iff (\omega -36{)}^2\ge 0,$$ it follows that $$\frac{(x+y+z+12{)}^2}{(x+y+z-6)}\ge 72.(3)$$ Equality holds when: $$\omega =x+y+z+12=36\iff x+y+z=24(4)$$ From relations (1) and (3) follows that: $$\frac{(x+y+z+12{)}^2}{(x+y+z-6)}=72,$$ and thus equations (2) are (4) are valid, and hence we have the system $$\left\{\begin{array}{l}(2\lambda -1)(x+y+z)=12(\lambda +1)\\ x+y+z=24\end{array}\right\}\Rightarrow \lambda =1.$$ For $\lambda =1$, from relations (2) we get the system : $$\left\{\begin{array}{l}y+z-x=4\\ x+z-y=8\\ x+y-z=12\end{array}\right\}\iff (x,y,z)=(10,8,6).$$ Therefore the unique solution of the problem is the triple $(x,y,z)=(10,8,6)$, taking in mind that it satisfies the equation $x+y+z=24$.