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
 A: 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 $x=y=z=\frac{1}{\sqrt{2}}$
A: You were very close, but you made a mistake.  You should have
$$\frac{1}{1+\tan^2 a}+\frac{1}{1+\tan^2b}\geq \frac{1}{\tan c}$$
instead.  Here is a similar approach.  
First, we assume that $x,y,z>0$. So, we can define $p,q,r\geq 0$ to be $\frac{\sqrt{1-x^2}}{x}$, $\frac{\sqrt{1-y^2}}{y}$, and $\frac{\sqrt{1-z^2}}{z}$, respectively.  The three inequalities become
$$\frac{1}{1+p^2}+\frac{1}{1+q^2}\geq \frac1r\wedge \frac{1}{1+q^2}+\frac{1}{1+r^2}\geq \frac1p \wedge \frac{1}{1+r^2}+\frac{1}{1+p^2}\geq \frac1q.$$
Adding all of these and then dividing the result by $2$ yield
$$\frac{1}{1+p^2}+\frac{1}{1+q^2}+\frac{1}{1+r^2}\geq \frac{1}{2p}+\frac{1}{2q}+\frac{1}{2r}.\tag{1}$$
However, by AM-GM, $1+p^2\ge 2p$, $1+q^2\ge 2q$, and $1+r^2\ge 2r$.  That is,
$$\frac{1}{1+p^2}+\frac{1}{1+q^2}+\frac{1}{1+r^2}\leq \frac{1}{2p}+\frac{1}{2q}+\frac{1}{2r}.\tag{2}$$
From (1) and (2), we must have
$$\frac{1}{1+p^2}+\frac{1}{1+q^2}+\frac{1}{1+r^2}= \frac{1}{2p}+\frac{1}{2q}+\frac{1}{2r},$$
which implies $1+p^2=2p$, $1+q^2=2q$, $1+r^2=2r$, so $p=q=r=1$ and $x=y=z=\frac1{\sqrt{2}}$.
Now, WLOG, if $x=0$, then $y^2\sqrt{1-z^2}\geq z$ and $z^2\sqrt{1-y^2}\geq y$.  Multiplying the two inequalities gives
$$y^2z^2\sqrt{1-y^2}\sqrt{1-z^2}\geq yz.$$
If $yz\neq 0$, then dividing by $yz$, we have
$$yz\sqrt{1-y^2}\sqrt{1-z^2}\geq 1.$$
But by AM-GM, $y\sqrt{1-y^2}=\sqrt{y^2(1-y^2)}\leq \frac{y^2+(1-y^2)}{2}=\frac12$ and similarly, $z\sqrt{1-z^2}\leq \frac12$.  So,
$$yz\sqrt{1-y^2}\sqrt{1-z^2}\leq \frac12\cdot\frac12=\frac14<1.$$
This is a contradiction, so $yz=0$, so $y=0$ or $z=0$, so $x=y=z=0$.  Therefore, if any of the variables is $0$, all of them are $0$.  There are then two solutions $$x=y=z=0 \wedge x=y=z=1/\sqrt{2}.$$
A: 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.
