If $A + A^t = 2I$ then $\det(A) \geq 1$

Note that $A - I$ is a real skew-symmetric matrix, so all of its eigenvalues are pure imaginary (or zero) and appear in complex conjugate pairs. Therefore the eigenvalues of $A$ are either $1$ or of the form $1 + c i, 1-c i$ for some $c\in\mathbb{R}$. Note that $(1+ci)(1-ci) = 1+c^2 \geq 1$. Therefore

$$\det(A) = \prod (1 + c^2) \geq 1$$

Let $A = I+B.$ One can immediately notice that $B$ is skew-symmetric or $B = -B^T.$ This means the eigenvalues of $A$ are precisely $1+\lambda_j$, where $\lambda_j$ are the eigenvalues of $B.$ Now, it's well-known that the eigenvalues of a skew-symmetric matrix comes in pairs $\pm i\lambda_j.$ This means that: $$\det A = \prod(1+\lambda^2_j)\geq 1.$$ You are encouraged to fill in the details separating the cases where $n$ is odd or even.