<s>[INST] 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
 [/INST] The hint:
$$(x^2+y^2)\sqrt{1-z^2}\geq z$$ it's
$$(x^2+y^2)^2(1-z^2)\geq z^2$$ or
$$(x^2+y^2)^2\geq((x^2+y^2)^2+1)z^2,$$ which gives
$$z^2\leq\frac{(x^2+y^2)^2}{(x^2+y^2)^2+1}\leq\frac{(x^2+y^2)^2}{2(x^2+y^2)}=\frac{x^2+y^2}{2}.$$
By the same way $$y^2\leq\frac{x^2+z^2}{2}$$ and $$x^2\leq\frac{y^2+z^2}{2},$$ which gives
$$x=y=z$$ and all inequalities the are equalities.
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