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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

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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 $x=y=z=\frac{1}{\sqrt{2}}$

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  • $\begingroup$ Also there is the case $x=0$, which gives $x=y=z=0$. Nice! +1. $\endgroup$ – Michael Rozenberg Nov 23 '18 at 21:27
  • $\begingroup$ but I wrotte $]0;1[$ , sooo we can't have $0$ as a solution. $\endgroup$ – user600785 Nov 23 '18 at 21:56
  • $\begingroup$ But by the given it gives a solution. You can't change the given.It's not fair. $\endgroup$ – Michael Rozenberg Nov 23 '18 at 22:07
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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 $$x=y=z=0 \wedge x=y=z=1/\sqrt{2}.$$

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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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  • $\begingroup$ why it gives $z^2\leq\frac{(x^2+y^2)^2}{(x^2+y^2)^2+1}\leq\frac{(x^2+y^2)^2}{2(x^2+y^2)}=\frac{x^2+y^2}{2}.$ $\endgroup$ – user600785 Nov 24 '18 at 10:16
  • $\begingroup$ @user600785 I used AM-GM: $(x^2+y^2)^2+1\geq2\sqrt{(x^2+y^2)^2\cdot1}=2(x^2+y^2).$ See also my post, I added something. $\endgroup$ – Michael Rozenberg Nov 24 '18 at 12:20

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