Solve in positive integers the equation $$\dfrac{x-1}{y+1}+\dfrac{y-1}{z+1}+\dfrac{z-1}{x+1}=1.$$
Proposed by Titu Andreescu, University of Texas at Dallas
Solution:
By cyclic permutation, we can assume $x \leq y \leq z$. Then, $$\dfrac{x-1}{x+1} \leq \dfrac{y-1}{y+1} \leq \dfrac{z-1}{z+1}.$$ By the AM-GM Inequality, we have $$1=\dfrac{x-1}{y+1}+\dfrac{y-1}{z+1}+\dfrac{z-1}{x+1} \geq 3\sqrt[3]{\dfrac{x-1}{x+1}\cdot \dfrac{y-1}{y+1}\cdot \dfrac{z-1}{z+1}} \geq 3\cdot\dfrac{x-1}{x+1},$$ which gives $x \leq 2$. We have two cases.
(i) $x=1$. Hence, $$\dfrac{y-1}{z+1}+\dfrac{z-1}{2}=1,$$ i.e. $$2y=-z^2+2z+5.$$ Then $-z^2+2z+5=2y \geq 2$, which gives $z^2-2z-3 \leq 0$, i.e. $(z-3)(z+1) \leq 0$. So, $z \leq 3$. We get the unique solution $y=1, z=3$.
(ii) $x=2$. Hence, $$\dfrac{1}{y+1}+\dfrac{y-1}{z+1}+\dfrac{z-1}{3}=1.$$ Since $y \geq 2$, then $$1>\dfrac{2}{z+1}+\dfrac{z-1}{3}.$$ Clearly, $z \leq 3$. If $z=2$, we get $(x,y,z)=(2,2,2)$. If $z=3$, we get a contradiction.
In conclusion, $(x,y,z) \in \{(1,1,3),(1,3,1),(3,1,1),(2,2,2)\}$.
By cyclic permutation, we can assume $x \leq y \leq z$. Then, $$\dfrac{x-1}{x+1} \leq \dfrac{y-1}{y+1} \leq \dfrac{z-1}{z+1}.$$ By the AM-GM Inequality, we have $$1=\dfrac{x-1}{y+1}+\dfrac{y-1}{z+1}+\dfrac{z-1}{x+1} \geq 3\sqrt[3]{\dfrac{x-1}{x+1}\cdot \dfrac{y-1}{y+1}\cdot \dfrac{z-1}{z+1}} \geq 3\cdot\dfrac{x-1}{x+1},$$ which gives $x \leq 2$. We have two cases.
(i) $x=1$. Hence, $$\dfrac{y-1}{z+1}+\dfrac{z-1}{2}=1,$$ i.e. $$2y=-z^2+2z+5.$$ Then $-z^2+2z+5=2y \geq 2$, which gives $z^2-2z-3 \leq 0$, i.e. $(z-3)(z+1) \leq 0$. So, $z \leq 3$. We get the unique solution $y=1, z=3$.
(ii) $x=2$. Hence, $$\dfrac{1}{y+1}+\dfrac{y-1}{z+1}+\dfrac{z-1}{3}=1.$$ Since $y \geq 2$, then $$1>\dfrac{2}{z+1}+\dfrac{z-1}{3}.$$ Clearly, $z \leq 3$. If $z=2$, we get $(x,y,z)=(2,2,2)$. If $z=3$, we get a contradiction.
In conclusion, $(x,y,z) \in \{(1,1,3),(1,3,1),(3,1,1),(2,2,2)\}$.
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