A geometry problem by A Former Brilliant Member

Since $\sec^2 \theta \in \left[1, \infty \right)$ , therefore we must have,

$\dfrac{4xy}{{\left(x+y\right)}^2} \ge 1 \\ \implies \dfrac{4xy}{{\left(x+y\right)}^2} - 1 \ge 0 \\ \implies \dfrac{4xy - {\left(x+y\right)}^2}{{\left(x+y\right)}^2} \ge 0 \\ \implies 4xy - {\left(x+y\right)}^2 \ge 0 \\ \implies {\left(x+y\right)}^2 - 4xy \le 0 \\ \implies {\left(x-y\right)}^2 \le 0 \\ \implies {\left(x-y\right)}^2 = 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \small\color{#3D99F6}{\text{As } x \text{ and } y \text{ are real. Therefore the square of their difference cannot be less than zero}} \\ \implies x=y$

Which is the only required condition for $\sec^2 \theta = \dfrac{4xy}{{\left(x+y\right)}^2}$ to be true for real numbers $x$ and $y$ .

If x=y=1, then the RHS equals 1 and since sec(0)=1, it is possible for the two sides to be equivalent.