Smallest eigenvalues of Sum of Two Positive Matrices Let $C = A + B$, where $A$, $B$, and $C$ are positive definite matrices. In addition, $C$ is fixed. Let $\lambda (A)$, $\lambda (B)$, and $\lambda (C)$ be smallest eigenvalues of $A$, $B$, and $C$, respectively. Is there any result about the smallest eigenvalues of $C$ in comparison with the sum of smallest eigenvalues of $A$ and $B$? Is it true that : $\lambda (A)$ + $\lambda (B)$ < $\lambda (C)$ ?
Moreover, what is the smallest possible value of $\lambda (A)$ + $\lambda (B)$ given a fixed $C$, and under what condition does this happen? 
Many thanks!
Xuan
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Question about $\lambda_{min} (A+B) > \lambda_{min} (A) + \lambda_{min} (B) $ can be seen from Weyl's inequality.
The remaining question is about the smallest attainable value of $\lambda_{min} (A) + \lambda_{min} (B) $ given a fixed $C$?
 A: If $S$ is a symmetric real matrix, then the minimal eigenvalue of $S$, $\lambda_{\min}(S)$ is given by 
$$\lambda_{\min}(S)=\min_{x\neq 0}\frac{x^tSx}{x^tx}.$$
Indeed, it's not hard to see that when $S$ is diagonal, and in the general case diagonalize $S$ in an orthonormal basis. 
By the properties of $\min$, this gives $\lambda_{\min}(A+B)\geq \lambda_{\min}(A)+\lambda_{\min}(B)$. Moreover, if we have equality $\frac{x^tSx}{x^tx}$, it's for an eigenvector for $\lambda_{\min}$, hence we have equality if $A$ and $B$ have a common eigenvector. 
A: I answer the second part of the question. For $n=1$ the answer is $\lambda_{min}(C)$. For $n > 1$, the answer is, $\lambda_{min}(A) + \lambda_{min}(B)$ can be arbitrarily close to $0$.
Since $C$ is a symmetric positive definite matrix, we consider the singular value decomposition of $C$.
$$
C  = \sum_{i=1}^n \lambda_i v_i v_i^T,$$
where $0 < \lambda_1 \leq \dots \leq \lambda_n$ are eigenvalues of $C$, and $v_1, \dots, v_n$ is an orthonormal basis. For sufficiently small $\varepsilon > 0$, let
\begin{align*}
A &= (1-\varepsilon) \lambda_1 v_1 v_1^T + \varepsilon\lambda_2 v_2 v_2^T + \frac{1}{2}\sum_{i=3}^n \lambda_i v_i v_i^T,\\
B &= \varepsilon \lambda_1 v_1 v_1^T + (1-\varepsilon)\lambda_2 v_2 v_2^T + \frac{1}{2}\sum_{i=3}^n \lambda_i v_i v_i^T.
\end{align*}
Now $A+B=C$, $A$ and $B$ are positive definite, and
$$\lambda_{min}(A) + \lambda_{min}(B) \leq 2\varepsilon,$$
 which can be arbitrarily close to $0$.
