contest-mathinequality
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
Let $f(x) = \sqrt{1-x^{2}}-p x-q$, then the condition becomes $$|f(x)| \leq \frac{\sqrt{2}-1}{2}$$ Thus, the extrema of $f$ should lie in $\left[ -\frac{\sqrt{2}-1}{2}, \frac{\sqrt{2}-1}{2}\right]$. Now, let's see how its extrema look like: $$f'(x) = \frac{-x}{\sqrt{1-x^2}} - p = 0 $$ After bringing to a common denominator, numerator should be equal to zero: $$ -x - p\sqrt{1-x^2} = 0 \\[2mm] x = r\sqrt{1-x^2}, \ \ (r=-p) \\[2mm] x^2 = r^2(1-x^2) \\[2mm] (1+r^2)x^2 = r^2 \\[2mm] x_0 = \pm \frac{r}{1+r^2} = \mp\frac{p}{\sqrt{1+p^2}}$$ However, since $x \ge 0$, we get $$x_0 = -\frac{p}{\sqrt{1+p^2}} $$ Therefore, $$\begin{align}f(x_0) &= \sqrt{1-\frac{p^2}{1+p^2}} + \frac{p^2}{\sqrt{1+p^2}}-q \\[1mm]&= \frac{1 + p^2}{\sqrt{1+p^2}} -q \\[1mm]&= \sqrt{1+p^2}-q \end{align}$$ Hence, the condition seems to be satisfied for all $p$ and $q$ such that $$\left| \sqrt{1+p^2} -q \right| \le \frac{\sqrt2 - 1}{2} \tag{1}$$ Also, we need to see the edge cases when $x = 0$ and $x = 1$, which give $$|f(0)| = \left|1 - q\right| \le \frac{\sqrt{2}-1}{2} \tag{2}$$ and $$|f(1)| = \left|-p - q\right| \le \frac{\sqrt{2}-1}{2} \tag{3}$$ Finally, combining $(1)$, $(2)$ and $(3)$ we get the final answer.
I didn't solve the last three inequalities putting all together, but after graphing I noticed (but still needs to be justified rigorously by solving the inequalities) that only one pair satisfies the statement, namely, $$(p,q) = \left(-1, \frac{\sqrt 2 + 1}{2}\right)$$
Below is the graph with those $p$ and $q$:
Best Answer
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 $x=y=z=\frac{1}{\sqrt{2}}$