falantrng wrote: Find all the real number triples $(x, y, z)$ satisfying the following system of inequalities under the condition $0 < x, y, z < \sqrt{2}$: $$y\sqrt{4-x^2y^2}\ge \frac{2}{\sqrt{xz}}$$$$x\sqrt{4-x^2z^2}\ge \frac{2}{\sqrt{yz}}$$$$z\sqrt{4-y^2z^2}\ge \frac{2}{\sqrt{xy}}$$. By AM-GM $$8\leq\sqrt{\prod_{cyc}\left(y^2xz\cdot(4-x^2y^2)\right)}=\sqrt{\prod_{cyc}\left(x^2y^2\cdot(4-x^2y^2)\right)}\leq\sqrt{\prod_{cyc}\left(\frac{x^2y^2+4-x^2y^2}{2}\right)^2}=8,$$which says $$x^2y^2=x^2z^2=y^2z^2=2$$or $$x=y=z=\sqrt[4]2.$$

My solution at the exam was using $\sqrt{4-x^2y^2}\le \frac{2}{xy}$. The rest is clear when you get: $z\ge x$, $x\ge y$ and $y\ge z$. Means $x=y=z$ so putting this to any of this three inequality will give us $x=y=z=\sqrt{\sqrt{2}}$

I analyzed the tie situation at the last moment and won 2 points