Is $\det(U_1\Lambda_1 U_1^t +U_2\Lambda_2 U_2^t +I)\le \det(\Lambda_1 +\Lambda_2 +I)$ correct? I want to simplify or find an upper bound for the  determinant 
$|K_1+K_2+I|$ where $I$ is identity matrix, $K_1$ and $K_2$ are positive semi-definite matrices of size $n$ and thus can be written as $K_1=U_1\Lambda_1 U_1^t$, $K_2=U_2\Lambda_2 U_2^t$,  in which  $U_1$ and $U_2$ are unitary matrices and $\Lambda_1, \Lambda_2$ are diagonal matrices. 
Can one say  $\det(U_1\Lambda_1 U_1^t +U_2\Lambda_2 U_2^t +I)\le \det(\Lambda_1  +\Lambda_2  +I)$?
Hint: when we have only $K_1$ (or $K_2$) we can write:
$|K_1+I|=|U_1\Lambda_1 U_1^t  +I|=|\Lambda_1 U_1^tU_1 +I|=|\Lambda_1   +I|$ 
where the second equality is due to the identity $\det(AB+I)=\det(BA+I)$.
 A: Unfortunately, the inequality you seek to prove  seems not to be established for the general positive semi-definite matrices. I can only prove it for very specific case of $K_1, K_2$  such that eigenvalues of the sum of these matrices are always less than or equal to the corresponding  sum  of eigenvalues of each of them. 
Below is the proof for this special case, if it is of any use for you. For more details see this article. 

Recall: 


*

*The set of positive semidefinite matrices is convex (see section "Further properties", property 12), i.e. $K_1, K_2$ are positive semidefinite $\implies  \alpha K_1 + (1-\alpha) K_2$ is also positive semidefinite for all $\alpha \in (0,1)$. By setting $\alpha  = \frac{1}{2}$, we establish that the sum of two positive semidefinite matrices $\left(K_1 + K_2\right) : = L
$ is also positive semidefinite.

*For two  positive semidefinite matrices  $A,B$  the following inequality folds: 
$\det(A+B)\ge \det(A) + \det(B)$. This is due to Minkowski's theorem 

I assume that by
$\det(U_1\Lambda_1 U_1^t +U_2\Lambda_2 U_2^t +I)\le \det(\Lambda_1 +\Lambda_2 +I)$ you mean
$$
\left| K_1  + K_2 + I \right|  \le \left| \Lambda_1 + \Lambda_2 + I \right|
$$
where $K_1=U_1\Lambda_1 U_1^t, K_2=U_2\Lambda_2 U_2^t$. 

Since $ L:=K_1 + K_2$ is positive semidefinite,  it can be diagonalized as $L = P D P^T$.  Then
$$
\left| K_1 + K_2 +I \right| = \left| L +I \right| = \left| D  + I \right| = \prod_{i=1}^n \big( \lambda_i\left( \right) + 1 \big),
$$
where $\lambda_1(Q), \dots, \lambda_n(Q)$  are the eigenvalues of a matrix  $Q$. 
On the other hand, 
$$
 \left| \Lambda_1 + \Lambda_2 + I \right| = \prod_{i=1}^n \big( \lambda_i\left( K_1\right)  +   \lambda_i\left( K_2\right)  +  1\big)
$$
If  for all $ i = 1, \dots,n$ we had the following inequality  $$
\lambda_i(L) =\lambda_i(K_1 + K_2) \le \lambda_i(K_1) +\lambda_i(K_2),
$$
it would  imply 
$$
\left| K_1 + K_2 +I \right| =  \prod_{i=1}^n \big( \lambda_i\left( L\right) + 1 \big)  
\le 
\prod_{i=1}^n \big( \lambda_i\left( K_1\right)  +   \lambda_i\left( K_2\right)  +  1\big) = \left| \Lambda_1 + \Lambda_2 + I \right|
$$
If the eigenvalues inequality does not hold, then the result would not be valid. 
