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Algebra Difficulty 7.1 National olympiad, round 2 Prove it

11 Prove: In the open interval (0,1)(0,1), there must exist four pairs of distinct positive numbers (a,b)(a≠b)(a, b)(a \neq b), satisfying:
(1−a2)(1−b2)>a2b+b2a−ab−18ab\sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}>\frac{a}{2 b}+\frac{b}{2 a}-a b-\frac{1}{8 a b}

Solution

11. Let a=cos⁡α,b=cos⁡β,α,β∈(0,π2)a=\cos \alpha, b=\cos \beta, \alpha, \beta \in\left(0, \frac{\pi}{2}\right), then
ab+(1−a2)(1−b2)=cos⁡(α−β)a b+\sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}=\cos (\alpha-\beta)

Square both sides, we have (1−a2)(1−b2)=12ab⋅[cos⁡2(α−β)−1]+a2b+b2a−ab\sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}=\frac{1}{2 a b} \cdot\left[\cos ^{2}(\alpha-\beta)-1\right]+\frac{a}{2 b}+\frac{b}{2 a}-a b.
When 0<α,β<π20<\alpha, \beta<\frac{\pi}{2} and 0<∣α−β∣<π60<|\alpha-\beta|<\frac{\pi}{6}, then
12ab⋅[cos⁡2(α−β)−1]>−18ab\frac{1}{2 a b} \cdot\left[\cos ^{2}(\alpha-\beta)-1\right]>-\frac{1}{8 a b}

The original inequality holds.
It is evident that within the open interval (0,π2)\left(0, \frac{\pi}{2}\right), it is possible to choose 4 pairs of distinct angle pairs (αi,βi)\left(\alpha_{i}, \beta_{i}\right), such that there exist two angle pairs (α,β)(\alpha, \beta), satisfying 0<∣α−β∣<π60<|\alpha-\beta|<\frac{\pi}{6}. Therefore, the conclusion holds.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.