Must They All Be Equal To Each Other?

here is another solution

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$x=2-\frac{2}{1+z^2}$ $y=2-\frac{2}{1+x^2}$ $z=2-\frac{2}{1+y^2}$ The minimum value of any of $x,y,z$ from the equations above ioccurs when the denominator is minimised, or when they are $0$ .

WLOG let $0 \leq z \leq y \leq x \leq 2$ . Note that the other case of $0 \leq z \leq x \leq y \leq 2$ can be handled similarly. $x \geq y \geq z$ $\Rightarrow -\frac{2}{1+z^2} \geq -\frac{2}{1+x^2} \geq -\frac{2}{1+y^2}$ $\Rightarrow 1+z^2 \geq 1+x^2 \geq 1+y^2$ $\Rightarrow z \geq y \geq x$ The last step followed since $x,y,z$ are all positive. Since $x \geq z$ and $z \leq x$ , we have $x=y=z$ $\Rightarrow x=\frac{2x^2}{1+x^2}$ Excluding the trivial solution, $1+x^2=2x$ $\Rightarrow (x-1)^2=0$ $\Rightarrow x=y=z=1$ Hence there is only $\boxed{1}$ solution.

Observe that, if any one of the variables is zero, it implies that all other variables are zero. Thus, without any loss of generality, we can focus on solutions in which none of the variables is zero. In this case, we can take reciprocal of the given three equations and sum them up to obtain $\frac{1}{x}+ \frac{1}{y}+\frac{1}{z}= \frac{3}{2} + \frac{1}{2}( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2})$ Rearranging the above, we obtain $(\frac{1}{x}-1)^2+ (\frac{1}{y}-1)^2+ (\frac{1}{z}-1)^2=0$ Which implies that the only other real solution is $x=y=z=1$

As the three equations are similar, we assume the solution to be $x=y=z=t$ . Then we have:

$\begin{aligned} \frac {2t^2}{1+t^2} & = t \\ \frac {2t^2}{1+t^2} - t & = 0 \\ t\left(\frac {2t}{1+t^2}-1\right) & = 0 \end{aligned}$

$\implies \begin{cases} t = 0 & \implies (x,y,z) = (0,0,0) \\ 2t - 1 - t^2 = 0 & \implies t = 1 \implies (x,y,z) = (1,1,1) \end{cases}$

Therefore, the number of solutions other than (0,0,0) is $\boxed{1}$ .

All I did was start by observing that $x,y,z \geq 0$ because the right hand sides of the equations must all be $\geq 0$ . Now assume all the variables are strictly positive (for obvious reasons). Then I observed that if $x=1$ , then the rest of the variables must equal 1 (and this is an if and only if relationship because of the symmetry involved in the three equations). Now I consider $x\neq 1$ , which I write equivalently as $(x-1)^2>0$ . This, in turn implies $x>y$ (obtained by adding $2x$ to both sides of the inequality, dividing by $1+x^2$ , multiplying by $x$ , and substituting $y$ for $\frac{2x^2}{1+x^2}$ ). However, if $x \neq 1$ , then $y\neq 1$ , and by similar reasoning, we can conclude $y>x$ . This is a contradiction. Therefore, there are no other solutions.