Jtdylktuy

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

Source : Test N°1 for IMO 2020 in Morocoo

The inequality is symetric.

so we can suppose that $xge y ge z$

the second inequality becomes
$$2x^2sqrt{1-x^2}ge x$$
$$2xsqrt{1-x^2}ge 1$$
$$4x^2-4x^4-1ge 0$$
$$-(2x^2-1)^2ge 0$$
$$2x^2-1=0$$

$$x=frac{1}{sqrt{2}}$$

By the same way , after remplacing $x$ by its value
we will find that $x=y=z=frac{1}{sqrt{2}}$

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 $p,q,rgeq 0$ 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
$$frac{1}{1+p^2}+frac{1}{1+q^2}geq frac1rwedge frac{1}{1+q^2}+frac{1}{1+r^2}geq frac1p wedge frac{1}{1+r^2}+frac{1}{1+p^2}geq frac1q.$$
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, $1+p^2ge 2p$, $1+q^2ge 2q$, and $1+r^2ge 2r$. 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 $y^2sqrt{1-z^2}geq z$ and $z^2sqrt{1-y^2}geq y$. Multiplying the two inequalities gives
$$y^2z^2sqrt{1-y^2}sqrt{1-z^2}geq yz.$$
If $yzneq 0$, then dividing by $yz$, we have
$$yzsqrt{1-y^2}sqrt{1-z^2}geq 1.$$
But by AM-GM, $ysqrt{1-y^2}=sqrt{y^2(1-y^2)}leq frac{y^2+(1-y^2)}{2}=frac12$ and similarly, $zsqrt{1-z^2}leq frac12$. So,
$$yzsqrt{1-y^2}sqrt{1-z^2}leq frac12cdotfrac12=frac14<1.$$
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 $$x=y=z=0 wedge x=y=z=1/sqrt{2}.$$

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
$$(x^2+y^2)^2geq((x^2+y^2)^2+1)z^2,$$ which gives
$$z^2leqfrac{(x^2+y^2)^2}{(x^2+y^2)^2+1}leqfrac{(x^2+y^2)^2}{2(x^2+y^2)}=frac{x^2+y^2}{2}.$$
By the same way $$y^2leqfrac{x^2+z^2}{2}$$ and $$x^2leqfrac{y^2+z^2}{2},$$ which gives
$$x=y=z$$ and all inequalities the are equalities.