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Tuesday, April 2, 2013

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.

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