Find the largest possible positive integer $n$, such that there exists $n$ distinct positive real numbers $x_1, x_2, \cdots, x_n $ satisfying the following inequality:
for any $ 1 \le i, j \le n$
$ (3x_i - x_j) (x_i - 3x_j) \ge (1 - x_ix_j) ^2$
Find largest pos int n with special condition
- Atonu Roy Chowdhury
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- Thanic Nur Samin
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Re: Find largest pos int n with special condition
Let $a_i=\tan ^{-1} x_i$.
Now, $(3x_i-x_j)(x_i-3x_j)\ge (1-x_ix_j)^2\Rightarrow \dfrac{x_i-x_j}{1+x_ix_j}\ge \dfrac{1}{\sqrt{3}}$ when $x_i> x_j$.
But this implies $\tan(a_i-a_j)\ge \dfrac{1}{\sqrt{3}}$ which means $a_i-a_j\ge 30^{\circ}$, whereas $0^{\circ}< a_k< 90^{\circ}$. If we take $n\ge 4$, then there must be two $a_k$'s so that their difference is at most $30^{\circ}$, so $n\le 3$. For $n=3$ we can easily find such a construction.
Now, $(3x_i-x_j)(x_i-3x_j)\ge (1-x_ix_j)^2\Rightarrow \dfrac{x_i-x_j}{1+x_ix_j}\ge \dfrac{1}{\sqrt{3}}$ when $x_i> x_j$.
But this implies $\tan(a_i-a_j)\ge \dfrac{1}{\sqrt{3}}$ which means $a_i-a_j\ge 30^{\circ}$, whereas $0^{\circ}< a_k< 90^{\circ}$. If we take $n\ge 4$, then there must be two $a_k$'s so that their difference is at most $30^{\circ}$, so $n\le 3$. For $n=3$ we can easily find such a construction.
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