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Friday, February 16, 2018

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).

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