59th Ukrainian National Mathematical Olympiad

Consider the following substitutions: $a=x+1>1$, $b=y+1>1$ and $c=z+1>1$. Then the first inequality is $ab\le c^2$. The second inequality can be rewritten the following way (using the first one): $$\begin{gathered} \frac{ab}{(a-1)(b-1)}\le {\left(\frac{c}{c-1}\right)}^2={\left(1+\frac{1}{c-1}\right)}^2\le {\left(1+\frac{1}{\sqrt{ab}-1}\right)}^2=\frac{ab}{(\sqrt{ab}-1{)}^2}\iff \\ (\sqrt{ab}-1{)}^2\le (a-1)(b-1)\iff a-2\sqrt{ab}+b\le 0\iff (\sqrt{a}-\sqrt{b}{)}^2\le 0\iff a=b, \end{gathered}$$ Moreover, equality is possible only if all the transitions satisfy the equality. So, the following equality has to hold: $ab=c^2\iff a=b=c\iff x=y=z$.

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Alternative solution.

Consider the following substitution: $x={\mathrm{tg}}^2\alpha$, $y={\mathrm{tg}}^2\beta$, $\alpha ,\beta \in (0,\frac{\pi }{2})$. Let the triple $(x,y,z)$ satisfy the conditions, then $$\{\begin{array}{l}\frac{1}{{\cos }^2\alpha {\cos }^2\beta }\le (z+1{)}^2,\\ \frac{1}{{\sin }^2\alpha {\sin }^2\beta }\le {\left(\frac{1}{z}+1\right)}^2,\\ 1\ge \left(\frac{1}{\cos \alpha \cos \beta }-1\right)\left(\frac{1}{\sin \alpha \sin \beta }-1\right)\\ \frac{1}{\cos \alpha \cos \beta }+\frac{1}{\sin \alpha \sin \beta }\ge \frac{1}{\cos \alpha \cos \beta \sin \alpha \sin \beta }\\ \sin \alpha \sin \beta +\cos \alpha \cos \beta =\cos (\alpha -\beta )\ge 1\Rightarrow \alpha =\beta \Rightarrow x=y.\end{array}\Rightarrow \{\begin{array}{l}z\ge \frac{1}{\cos \alpha \cos \beta }-1,\\ \frac{1}{z}\ge \frac{1}{\sin \alpha \sin \beta }-1,\\ 1\ge \left(\frac{1}{\cos \alpha \cos \beta }-1\right)\left(\frac{1}{\sin \alpha \sin \beta }-1\right)\\ \frac{1}{\cos \alpha \cos \beta }+\frac{1}{\sin \alpha \sin \beta }\ge \frac{1}{\cos \alpha \cos \beta \sin \alpha \sin \beta }\\ \sin \alpha \sin \beta +\cos \alpha \cos \beta =\cos (\alpha -\beta )\ge 1\Rightarrow \alpha =\beta \Rightarrow x=y.\end{array}$$

$$\{\begin{array}{l}(x+1{)}^2\le (z+1{)}^2,\\ {\left(\frac{1}{x}+1\right)}^2\le {\left(\frac{1}{z}+1\right)}^2,\end{array}\Rightarrow \{\begin{array}{l}x\le z,\\ \frac{1}{x}\le \frac{1}{z},\end{array}\Rightarrow \{\begin{array}{l}x\le z,\\ z\le x,\end{array}\Rightarrow x=z,$$

that leads to the solution.