Solve this eqtn (seems interesting)

by\ 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...

good work dimensions but we also have to include x=y=z=0

yup u both are rite.

After 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