1wow all i cud infer at d first ook is x y z all are positive hehe
\frac{4y^{2}}{4y^{2}+1}=z
\frac{4z^{2}}{4z^{2}+1}=x
\frac{4x^{2}}{4x^{2}+1}=y
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1by\ AM-GM\ inequality\ we\ have,\\ \\ (1+\frac{1}{4x^2}\geq \frac{1}{x}), (1+\frac{1}{4y^2}\geq \frac{1}{y}),(1+\frac{1}{4z^2}\geq \frac{1}{z})\\ \\ adding\ all\ three\\ \\ \frac{4x^2+1}{4x^2} +\frac{4y^2+1}{4y^2} + \frac{4z^2+1}{4z^2} \geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\\ \\ but\ equality\ is\ holding\ in\ the\ above\ inequation\ by\ given\\ conditions\ which\ is\ true\ when\ all\ variables\ are\ equal\\ \\ i.e\ \frac{1}{x}= \frac{1}{y}=\frac{1}{z},\\ \\ solving\ for\ them\ we\ get\\ \\ x=y=z=\frac{1}{2}
this must be the only possible solution...
341After noting that (0,0,0) is a solution, we note that if (x,y,z) is another solution then x,y,z>0
x = \frac{4z^2}{1+4z^2} \le \frac{4z^2}{4z} \le z
Thus, we have x \le z \le y \le x
from which we obtain x=y=z = 1/2