Maths Olympiad Prep

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Algebra Difficulty 6.0 AIME, harder Prove it Iran

Do there exist two functions f,g:RRf, g: \mathbb{R} \to \mathbb{R} such that for all xyx \neq y the following inequality holds?
f(x)f(y)+g(x)g(y)>1 |f(x) - f(y)| + |g(x) - g(y)| > 1

Solution

For all xyx \neq y we have:
(f(x)f(y))2+(g(x)g(y))212(f(x)f(y)+g(x)g(y))2>12 (f(x) - f(y))^2 + (g(x) - g(y))^2 \geq \frac{1}{2} (|f(x) - f(y)| + |g(x) - g(y)|)^2 > \frac{1}{2}
For every xx, consider the point (f(x),g(x))(f(x), g(x)) in the R2\mathbb{R}^2 plane. The above inequality shows that the distance between every two of these points is more than 12\frac{1}{\sqrt{2}}. For every point (f(x),g(x))(f(x), g(x)), consider a circle with center at this point and a radius of 122\frac{1}{2\sqrt{2}}, so not two of these circles intersect. On the other hand the number of these points are uncountable, so we have an uncountable number of circles in the plane that no two of them intersect and this is impossible; because there is a point with rational coordinates in every one of the circles, hence we would have an uncountable number of points with rational coordinates and since we know that the total number of the points with rational coordinates is countable, this is a contradiction; so no two functions with this property exist.

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