Mathematical Reflections 2015, Issue 3 - Problem J342

Problem:
Solve in positive real numbers the system of equations $$\dfrac{x^3}{4}+y^2+\dfrac{1}{z}=\dfrac{y^3}{4}+z^2+\dfrac{1}{x}=\dfrac{z^3}{4}+x^2+\dfrac{1}{y}=2.$$

Proposed by Titu Andreescu, University of Texas at Dallas

Solution:
Adding the three equations, we get $$\left(\dfrac{x^3}{4}+x^2+\dfrac{1}{x}-2\right)+\left(\dfrac{y^3}{4}+y^2+\dfrac{1}{y}-2\right)+\left(\dfrac{z^3}{4}+z^2+\dfrac{1}{z}-2\right)=0.$$
Let $f(t)=\dfrac{t^3}{4}+t^2+\dfrac{1}{t}-2$. We have
$$f(t)=\dfrac{t^4+4t^3-8t+4}{4t}=\dfrac{(t^2+2t-2)^2}{4t} \geq 0$$ for all $t \in \mathbb{R}^+$ and $f(t)=0$ if and only if $t^2+2t-2=0$, i.e. $t=\sqrt{3}-1$. It follows that $f(x)+f(y)+f(z)=0$ if and only if $x=y=z=\sqrt{3}-1$.