system of equations with three equations.

Question

We have to find all real solutions to this system of equations: $$x=\frac{4z^2}{1+4z^2},y=\frac{4x^2}{1+4x^2},z=\frac{4y^2}{1+4y^2}$$

Answer 1

$x=y=z=0$ and $x=y=z=\frac{1}{2}$ are the only real solutions.

Answer 2

Let us start solving by taking the first term:

$x=\frac{4z^2}{1+4z^2}$

Assuming $z$ not equal to $0$.

$\frac{x}{z} = \frac{4z}{1+4z^2}$---(1)

Since Arithmetic Mean >= Geometric Mean , we get:

$\frac{1+4z^2}{2} >= 2z$

$=> \frac{4z}{1+4z^2} <= 1$ ---(2)

From (2) and (1), we get :

$\frac{x}{z} <= 1$

Similarly for the other two terms we get :

$\frac{y}{x} <= 1$

$\frac{z}{y} <= 1$

This is possible only if : $x=y=z$

From this we get $x=y=z=1/2$

Now start from beginning by multiplying the 3 equations:

You get $xyz=0$, which implies one of the three is zero. Assuming $x=0$ and continuing you get the remaining terms to be zero as well.

Hence, $x=y=z=0$ is another solution.

Hence $(x,y,z) = (0,0,0) and (1/2,1/2,1/2)$