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Ukrainian National Mathematical Olympiad

Algebra Difficulty 6.2 National Olympiad Prove it Ukraine

Determine all positive x,y,zx, y, z, that satisfy the following system of inequalities:
{(x+1)(y+1)≤(z+1)2,(1x+1)(1y+1)≤(1z+1)2. \begin{cases} (x+1)(y+1) \le (z+1)^2, \\ \left(\frac{1}{x}+1\right)\left(\frac{1}{y}+1\right) \le \left(\frac{1}{z}+1\right)^2. \end{cases}

Solutions — 2

Solution 1

Consider the following substitutions: a=x+1>1a = x+1 > 1, b=y+1>1b = y+1 > 1 and c=z+1>1c = z+1 > 1. Then the first inequality is ab≤c2ab \le c^2. The second inequality can be rewritten the following way (using the first one):
ab(a−1)(b−1)≤(cc−1)2=(1+1c−1)2≤(1+1ab−1)2=ab(ab−1)2⇔(ab−1)2≤(a−1)(b−1)⇔a−2ab+b≤0⇔(a−b)2≤0⇔a=b, \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} \Leftrightarrow \\ (\sqrt{ab}-1)^2 \le (a-1)(b-1) \Leftrightarrow a - 2\sqrt{ab} + b \le 0 \Leftrightarrow (\sqrt{a}-\sqrt{b})^2 \le 0 \Leftrightarrow a = b,
Moreover, equality is possible only if all the transitions satisfy the equality. So, the following equality has to hold: ab=c2⇔a=b=c⇔x=y=zab = c^2 \Leftrightarrow a = b = c \Leftrightarrow x = y = z.

Solution 2

Consider the following substitution: x=tg⁡2αx = \operatorname{tg}^2 \alpha, y=tg⁡2βy = \operatorname{tg}^2 \beta, α,β∈(0,π2)\alpha, \beta \in (0, \frac{\pi}{2}). Let the triple (x,y,z)(x, y, z) satisfy the conditions, then
{1cos⁡2αcos⁡2β≤(z+1)2,1sin⁡2αsin⁡2β≤(1z+1)2,1≥(1cos⁡αcos⁡β−1)(1sin⁡αsin⁡β−1)1cos⁡αcos⁡β+1sin⁡αsin⁡β≥1cos⁡αcos⁡βsin⁡αsin⁡βsin⁡αsin⁡β+cos⁡αcos⁡β=cos⁡(α−β)≥1⇒α=β⇒x=y.⇒{z≥1cos⁡αcos⁡β−1,1z≥1sin⁡αsin⁡β−1,1≥(1cos⁡αcos⁡β−1)(1sin⁡αsin⁡β−1)1cos⁡αcos⁡β+1sin⁡αsin⁡β≥1cos⁡αcos⁡βsin⁡αsin⁡βsin⁡αsin⁡β+cos⁡αcos⁡β=cos⁡(α−β)≥1⇒α=β⇒x=y. \begin{cases} \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{cases} \Rightarrow \begin{cases} 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{cases}

{(x+1)2≤(z+1)2,(1x+1)2≤(1z+1)2,⇒{x≤z,1x≤1z,⇒{x≤z,z≤x,⇒x=z, \begin{cases} (x+1)^2 \le (z+1)^2, \\ \left(\frac{1}{x}+1\right)^2 \le \left(\frac{1}{z}+1\right)^2, \end{cases} \Rightarrow \begin{cases} x \le z, \\ \frac{1}{x} \le \frac{1}{z}, \end{cases} \Rightarrow \begin{cases} x \le z, \\ z \le x, \end{cases} \Rightarrow x = z,

that leads to the solution.

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