# 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

Source : Test N°1 for IMO 2020 in Morocoo

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

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

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 $$x=y=z=0 \wedge x=y=z=1/\sqrt{2}.$$

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.

• why it gives $z^2\leq\frac{(x^2+y^2)^2}{(x^2+y^2)^2+1}\leq\frac{(x^2+y^2)^2}{2(x^2+y^2)}=\frac{x^2+y^2}{2}.$ – user600785 Nov 24 '18 at 10:16
• @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. – Michael Rozenberg Nov 24 '18 at 12:20