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 $\sqrt 2 < 2\sqrt{8/5}$.

[edit] Since we are still there, we may ask what is the optimal configuration for four points in the square, that is, the one forming four triangles with larger minimal altitude $h$. Four points on a square either form a convex quadrilateral (A), or a triangle with an interior point (B). It is convenient to consider separately the two cases.

(A) Among all choices of four points that form a convex quadrilateral, by a symmetrization argument, the optimal configuration can be reached on rectangles (any quadrilateral contains a rectangle with the diagonals on the same lines, and with same minimal altitude $h$). Moreover, a rectangle can be enlarged and rotated untill all its vertices touch the edges of the unit square, and this will not decrease $h$. This way it follows that the optimal configuration for convex quadrilaterals is the unit square, giving the optimal $h_A=\sqrt{2}/2$.

(B) Given a triangle inside the unit square, the choice of a fourth point inside the triangle, giving the best $h$ is the incenter of the triangle (all three interior altitudes are equal). In this case, the problem may be stated: find the largest radius $h$ of a circle contained in a triangle contained in the unit square. This is a computation that you may like to complete (one vertex of the triangle must coincide with a vertex of the unit square, analogous argument as before); in any, case it will give an optimal $h_B < 1/2$, worse than the above $h_A$.

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 inside the unit square can have an altitude larger than $\sqrt 2 < 2\sqrt{8/5}$.

[edit] Since we are still here, we may ask what is the optimal configuration for four points in the square, that is, the one forming four triangles with larger minimal altitude $h$. Four points on a square either form a convex quadrilateral (A), or a triangle with an interior point (B). It is convenient to consider separately the two cases.

(A) Among all choices of four points that form a convex quadrilateral, by a symmetrization argument, the optimal configuration can be reached on rectangles (any quadrilateral contains a rectangle with the diagonals on the same lines, and with same minimal altitude $h$). Moreover, a rectangle can be enlarged and rotated until all its vertices touch the edges of the unit square, and this will not decrease $h$. This way it follows that the optimal configuration for convex quadrilaterals is the unit square, giving the optimal $h_A=\sqrt{2}/2$.

(B) Given a triangle inside the unit square, the choice of a fourth point inside the triangle, giving the best $h$ is the incenter of the triangle (all three interior altitudes are equal). In this case, the problem may be stated: find the largest radius $h$ of circles contained in triangles contained in the unit square. This is a computation that you may like to complete (one vertex of the triangle must coincide with a vertex of the unit square, by an analogous argument as before); in any, case it will give an optimal $h_B < 1/2$, worse than the above $h_A$.

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 $\sqrt 2 < 2\sqrt{8/5}$.

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 $\sqrt 2 < 2\sqrt{8/5}$.

[edit] Since we are still there, we may ask what is the optimal configuration for four points in the square, that is, the one forming four triangles with larger minimal altitude $h$. Four points on a square either form a convex quadrilateral (A), or a triangle with an interior point (B). It is convenient to consider separately the two cases.

(A) Among all choices of four points that form a convex quadrilateral, by a symmetrization argument, the optimal configuration can be reached on rectangles (any quadrilateral contains a rectangle with the diagonals on the same lines, and with same minimal altitude $h$). Moreover, a rectangle can be enlarged and rotated untill all its vertices touch the edges of the unit square, and this will not decrease $h$. This way it follows that the optimal configuration for convex quadrilaterals is the unit square, giving the optimal $h_A=\sqrt{2}/2$.

(B) Given a triangle inside the unit square, the choice of a fourth point inside the triangle, giving the best $h$ is the incenter of the triangle (all three interior altitudes are equal). In this case, the problem may be stated: find the largest radius $h$ of a circle contained in a triangle contained in the unit square. This is a computation that you may like to complete (one vertex of the triangle must coincide with a vertex of the unit square, analogous argument as before); in any, case it will give an optimal $h_B < 1/2$, worse than the above $h_A$.