Find all real solutions to the following system of equations.  Carefully justify your answer.
\[ \left\{ \begin{array}{c} \displaystyle\frac{4x^2}{1+4x^2} = y \\ \\ \displaystyle\frac{4y^2}{1+4y^2} = z \\ \\ \displaystyle\frac{4z^2}{1+4z^2} = x \end{array} \right. \]
1. We start with the given system of equations:
   \[
   \left\{ 
   \begin{array}{c} 
   \displaystyle\frac{4x^2}{1+4x^2} = y \\  
   \displaystyle\frac{4y^2}{1+4y^2} = z \\  
   \displaystyle\frac{4z^2}{1+4z^2} = x 
   \end{array} 
   \right.
   \]

2. Multiply all three equations together:
   \[
   \left(\frac{4x^2}{1+4x^2}\right) \left(\frac{4y^2}{1+4y^2}\right) \left(\frac{4z^2}{1+4z^2}\right) = y \cdot z \cdot x
   \]

3. Simplify the left-hand side:
   \[
   \frac{4x^2}{1+4x^2} \cdot \frac{4y^2}{1+4y^2} \cdot \frac{4z^2}{1+4z^2} = \frac{64x^2y^2z^2}{(1+4x^2)(1+4y^2)(1+4z^2)}
   \]

4. Equate the simplified left-hand side to the right-hand side:
   \[
   \frac{64x^2y^2z^2}{(1+4x^2)(1+4y^2)(1+4z^2)} = xyz
   \]

5. Assume none of \(x, y, z\) are zero. If one of \(x, y, z\) is zero, then we obtain the trivial solution \(\boxed{(x, y, z) = (0, 0, 0)}\).

6. Simplify the equation by dividing both sides by \(xyz\):
   \[
   \frac{64xyz}{(1+4x^2)(1+4y^2)(1+4z^2)} = 1
   \]

7. Rearrange the equation:
   \[
   (1+4x^2)(1+4y^2)(1+4z^2) = 64xyz
   \]

8. Consider the function \(f(t) = \frac{4t^2}{1+4t^2}\). Note that \(f(t)\) is an increasing function for \(t \geq 0\) and \(f(t) \leq 1\).

9. By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
   \[
   \frac{1}{x} + 4x \geq 4 \quad \text{with equality if and only if} \quad x = \frac{1}{2}
   \]

10. Apply the AM-GM inequality to each term:
    \[
    \frac{1}{x} + 4x \geq 4, \quad \frac{1}{y} + 4y \geq 4, \quad \frac{1}{z} + 4z \geq 4
    \]

11. Since equality holds if and only if \(x = y = z = \frac{1}{2}\), we conclude:
    \[
    x = y = z = \frac{1}{2}
    \]

12. Verify the solution by substituting \(x = y = z = \frac{1}{2}\) back into the original equations:
    \[
    \frac{4 \left(\frac{1}{2}\right)^2}{1 + 4 \left(\frac{1}{2}\right)^2} = \frac{1}{2}, \quad \frac{4 \left(\frac{1}{2}\right)^2}{1 + 4 \left(\frac{1}{2}\right)^2} = \frac{1}{2}, \quad \frac{4 \left(\frac{1}{2}\right)^2}{1 + 4 \left(\frac{1}{2}\right)^2} = \frac{1}{2}
    \]

    All equations are satisfied, confirming the solution.