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The quantity at the numerator is the length of the largest edge of the triangle with vertices $(x_i,y_i), (x_j,y_j), (x_k,y_k)$, squared. The denominator is its area, squared, so the fraction is four times the reciprocal squared of the minimal altitude of that triangle. The whole question may be therefore rephrased more geometrically: Are there $4$ points in the unit square such that the $12$ altitudes of the $4$ triangles they form are all larger than $2\sqrt{8/5}$? No, because no triangle iside the unit square can have an altitude larger than $\sqrt 2 < 2\sqrt{8/5}$.