bound on eigenvalues of sum of positive-semidefinites matrix

I have 2 positive-semidefinites matrix A and B.

I know that A+B is also a positive-semidefinites matrix.

I need to prove that the maximal eigenvalue of A+B is bounded by the sum of the maximal eigen value of A and the maximal eigenvalue of B.

Any ideas?

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Take the 2-norm of either of your matrices. Since they are symmetric, the 2-norm equals the absolute value of the largest eigenvalue. What inequality do norms satisfy? – Alex R. Dec 19 '12 at 21:44
@Alex The question doesn't say the matrices are symmetric. It may be reasonable to infer that they are from them being semi definite. By definition, you can have non-symmetric semidefinite matrices $A$ such that $x^TAx\geq0$. They do not induce a norm, however. – Daryl Dec 19 '12 at 22:25
@Daryl: Consider $A':=\frac{1}{2}(A+A^T)$ and similarly for $B'$. Notice that $A',B'$ are now symmetric. Consider the Rayleigh quotient of $A$ and $A'$. What can you say about the relation between eigenvalues? – Alex R. Dec 19 '12 at 22:50
@Alex I wasn't saying you were incorrect. The step that you took to assume symmetry may not have been so obvious. To answer your question, they are clearly equal. – Daryl Dec 19 '12 at 23:16
@Daryl: I'm not so sure they are equal. It's true that the eigenvalues of $A,A^T$ are equal but when summed as above the relationship is not necessarily preserved (particularly when $A$ is not symmetric to begin with). Alternately, if you deal with the Rayleigh quotient directly (not doing this symmetrization business) the result should also follow. – Alex R. Dec 19 '12 at 23:19

First, if the matrices are not symmetric it is not true. If $A=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$, then the maximum eigenvalue of $A$ is $1$ and the eigenvalues of $\frac{1}{2}(A+A^T)$ are $\frac{1}{2},\frac{3}{2}$ (hence $A$ is positive definite).
Now set $B = A^T$. Then the maximum eigenvalue of $A+B=2 \frac{1}{2}(A+A^T)$ is $3$, but the sum of the maximum eigenvalues is $2$.
However, if the matrices are symmetric, then it is clear from the diagonal form that $\lambda_{\max}(A) = \|A\|$, and since $\|A+B\|\leq \|A\|+\|B\|$, you have $\lambda_{\max}(A+B) \leq \lambda_{\max}(A) + \lambda_{\max}(B)$.