Mathematical Reflections 2013, Issue 1 - Problem S253

Problem:
Solve in positive real numbers the system of equations:
$$\begin{array}{rcl}(2x)^{2013}+(2y)^{2013}+(2z)^{2013} & = & 3 \\ xy+yz+zx+2xyz & = & 1 \end{array}$$

Proposed by Roberto Bosch Cabrera.

Solution:
Let $2x=a, 2y=b, 2z=c$. Then, the given system of equations is equivalent to
$$\begin{array}{rcl}a^{2013}+b^{2013}+c^{2013} & = & 3 \\ ab+bc+ca+abc & = & 4. \end{array}$$ Using the AM-GM Inequality in the first equation, we get $abc \leq 1$ and from the second equation we get $ab+bc+ca \geq 3$. Suppose without loss of generality that $a \leq b \leq c$. By Chebyshev's Inequality, we have
$$\begin{equation} 1 \leq \dfrac{ab+bc+ca}{3} \leq \dfrac{(a+b+c)^2}{9},                           (1) \end{equation}$$
therefore $a+b+c \geq 3$. Using Chebyshev's Inequality once again, we have $$\dfrac{a^{n-1}+b^{n-1}+c^{n-1}}{3} \leq \left(\dfrac{a+b+c}{3}\right)\left(\dfrac{a^{n-1}+b^{n-1}+c^{n-1}}{3}\right) \leq \dfrac{a^n+b^n+c^n}{3}$$ for all $n \in \mathbb{N}^*$. Since $\dfrac{a^{2013}+b^{2013}+c^{2013}}{3}=1$, we get $$a^n+b^n+c^n \leq 3$$ for all positive integers $n < 2013$, and in particular $a+b+c \leq 3$. This gives $a+b+c=3$, and from (1) we get $ab+bc+ca=3$, which implies $abc=1$. Therefore, $a+b+c=3\sqrt[3]{abc}$, so $a=b=c=1$, i.e. $x=y=z=1/2$ is the only solution to the given system of equations.