Gazeta Matematica 2/2017, Problem 27335

Problem:
Find all real solutions $(x,y,z)$ with $x,y,z \in \left(\dfrac{3}{2},+\infty\right)$ to the equation
$$\dfrac{(x+1)^2}{y+z-1}+\dfrac{(y+2)^2}{z+x-2}+\dfrac{(z+3)^2}{x+y-3}=18.$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
By the Cauchy-Schwarz Inequality, we have $$\left((y+z-1)+(z+x-2)+(x+y-3)\right)\cdot18 \geq \left((x+1)+(y+2)+(z+3)\right)^2,$$ i.e. $$(x+y+z+6)^2 \leq 36(x+y+z-3).$$ Let $t=x+y+z+6$. We have $$t^2 \leq 36(t-9),$$ i.e. $$(t-18)^2 \leq 0.$$ It follows that $t=18$, i.e.
\begin{equation}\label{first-eq}
x+y+z=12.
\end{equation}
The equality is attained when $$\dfrac{x+1}{y+z-1}=\dfrac{y+2}{z+x-2}=\dfrac{z+3}{y+z-3}=\lambda, \qquad \lambda \in \mathbb{R}.$$ So, $$\begin{array}{lll} \lambda(y+z)-x&=&\lambda+1 \\ \lambda(z+x)-y&=&2(\lambda+1) \\ \lambda(x+y)-z&=&3(\lambda+1). \end{array}$$ Adding side by side the three equations, we get $(2\lambda-1)(x+y+z)=6(\lambda+1)$ and by \eqref{first-eq}, we obtain $\lambda=1$. So,
$$\begin{array}{lll} y+z-x&=&2 \\ z+x-y&=&4 \\ x+y-z&=&6. \end{array}$$ Comparing these three equations with \eqref{first-eq}, we get $(x,y,z)=(5,4,3)$.