$\det(S) \le det(A+S)$ $A$ is real skew symmetric matrix
$S$ is a positive-definite symmetric matrix
Prove that $\det(S) \le \det(A+S)$
As $S$ is diagonalizable, we can reduce the problem to :
  for any real skew symmetric matrix $A$ and any diagonal matrix D with positive entries, prove that $\det(S) \le \det(D+S)$
I know that $A$ has only $0$ as real eigenvalue or imaginary numbers (conjugate each other). So $\det(A) \ge 0$. 
But i don't see how to calculate $\det(D+S)$ ?
Any hint ?
Merry Christmas !
 A: Hint: You made a mistake and you should try to prove $\det(D)\leq \det(D+A)$.  You may further assume that $D$ is the identity matrix.  Using the fact that the eigenvalues of $A$ are $0$ or purely imaginary complex numbers that come in conjugate pairs, the claim should be now trivial.
A: Let $A$ a real skew symmetric matrix and $S$ a positive-definite symmetric matrix
1) The eigenvalues of $A$ are purely imaginary complex numbers that come in conjugate pairs
2) As $S$ is positive-definite symmetric, we can find a positive-definite symmetric $T$ such that $$S=T^2.$$
3) we have $$\det(A+S) = \det(A+T^2) = \det(T(T^{-1}AT^{-1}+I_n)T) = \det(T^2)\det(\underbrace{T^{-1}AT^{-1}}_{=B}+I_n)$$ and $B$ is skew-symmetric
We want to prove $\det(I+B) \ge 1$.
4) The eigenvalues of $I+B$ are $1+i\alpha_k$. They are conjugate each other 
($\alpha_k$ could be zero).
Let $p$ the number of $\alpha_k>0$ ($p$ could be zero).
$$\det(I+B) = \prod_{k=1}^p(1+i\alpha_k)\prod_{k=1}^p(1-i\alpha_k) = \prod_{k=1}^p(1+\alpha_k^2) \ge 1.$$
We have proved that $\det(I+B) \ge 1$ which implies $\det(A+S) \ge \det(S)$. 
