How prove this matrix eigenvalue inequality $\lambda_{3}+\lambda_{2}>\lambda_{1}$ Question:
let $A_{n\times n} $ is Real symmetric postive matrices
 ,and  let $\lambda_{1},\lambda_{2},\lambda_{3}$ is $A_{n\times n},P_{r\times r},R_{s\times s}$ largest eigenvalue,  respectively.and such
$$A=\begin{bmatrix}
P&Q\\
Q^T&R
\end{bmatrix}$$
 show that:
$$\lambda_{3}+\lambda_{2}>\lambda_{1}$$
my try:without loss of we let
$$r\ge s$$
maybe follow use  Singular value decomposition？
then I can't,Thank you 
 A: Let $u \in M_{n\times 1}(\mathbb{R})$ be a $n \times 1$ unit column vector that maximizes $u^\top A u$. Rewrite it
as 
$$u = \begin{pmatrix}\alpha u_2\\ \beta u_3\end{pmatrix}$$
where $\alpha, \beta \in \mathbb{R}$, $u_2 \in M_{r\times 1}(\mathbb{R})$ and $u_3 \in M_{s\times 1}(\mathbb{R})$ are unit $r\times 1$ and $s \times 1$ column vectors. By definition, we have $\alpha^2 + \beta^2 = 1$. Let $v \in M_{n\times 1}(\mathbb{R})$ be 
another $n \times 1$ column vector defined by:
$$v = \begin{pmatrix}\beta u_2\\-\alpha u_3\end{pmatrix}$$
It is easy to see $v$ is also an unit column vector. Since $A$ is positive definite, we have:
$$\begin{align}
\lambda_1 =  u^\top A u 
< & u^\top A u + v^\top A v\\ 
= & \begin{pmatrix}\alpha u_2\\ \beta u_3\end{pmatrix}^\top
\begin{pmatrix}P & Q\\Q^\top & R\end{pmatrix}
\begin{pmatrix}\alpha u_2\\ \beta u_3\end{pmatrix}
+ \begin{pmatrix}\beta u_2\\ -\alpha u_3\end{pmatrix}^\top
\begin{pmatrix}P & Q\\Q^\top & R\end{pmatrix}
\begin{pmatrix}\beta u_2\\ -\alpha u_3\end{pmatrix}\\
= & (\alpha^2 + \beta^2) (u_2^\top P u_2 + u_3^\top R u_3)\\
\le & \lambda_2 u_2^\top u_2 + \lambda_3 u_3^\top u_3\\
= & \lambda_2 + \lambda_3
\end{align}$$
