52. On the plane, a square $A B C D$ is given. Find the minimum of the ratio $\frac{O A+O C}{O B+O D}$, where $O$ is an arbitrary point on the plane.
93.52. We will prove that $\sqrt{2(OA + OC)} \geqslant OB + OD$. Squaring the inequality and eliminating the denominator, since the equality $OA^2 + OC^2 = OB^2 + OD^2$ is a well-known fact, we get

$$
2[OA^2 + OC^2 + 2OA \cdot OC] \geqslant [OB^2 + OD^2 + 2OB \cdot OD]
$$

or

$$
OB^2 + OD^2 + 4OA \cdot OC \geqslant 2OB \cdot OD
$$

which is obvious, as the sum of squares is not less than twice the product. Finally, by choosing $O=A$, we get that the value $1 / \sqrt{2}$ is achieved, and therefore the minimum is $1 / \sqrt{2}$.