Find all real numbers $x,y,z\in [0,1]^3$ such that $(x^2+y^2)\sqrt{1-z^2}\ge z$.... Such that:
$$(x^2+y^2)\sqrt{1-z^2}\ge z$$
and
$$(z^2+y^2)\sqrt{1-x^2}\ge x$$
and
$$(x^2+z^2)\sqrt{1-y^2}\ge y$$
Since $x,y,z$ $\in ]0,1[^3$
then , there are some real numbers $a,b,c$ such that
$\cos a=x, \cos b=y , \cos c=z$
After some manipulations , we find that :
$$\frac{1}{1+\tan^2 a}+\frac{1}{1+\tan^2 b}\ge \frac{1}{\tan c}$$
.... same for other inequalities
I don't know what i must do now

The inequality is symetric.
so we can suppose that $x\ge y \ge z$ 
the second inequality becomes
$$2x^2\sqrt{1-x^2}\ge x$$
$$2x\sqrt{1-x^2}\ge 1$$
$$4x^2-4x^4-1\ge 0$$
$$-(2x^2-1)^2\ge 0$$
$$2x^2-1=0$$
$$x=\frac{1}{\sqrt{2}}$$
By the same way , after remplacing $x$ by its value
we will find that $x=y=z=\frac{1}{\sqrt{2}}$
