Max eigenvalue of symmetric matrix and its relation to diagonal values I saw few questions about it, but still can't understand. 
Let $A$ be a symmetric matrix and $\lambda_{\max}$ its largest eigenvalue. Is the following true for all $A$?
$$
\lambda_{\max} \ge a_{ii} \forall i
$$
That is, is the largest eigenvalue of a symmetric matrix always greater than any of its diagonal entries?
Is it somehow related to spectral radius and the following equation?
$$
\rho(A)=\max|\lambda_i|.
$$

Let $\lambda_1,\cdots, \lambda_n$ be eigenvalues of $A $ in increasing order,i.e. $\lambda_n$ is the maximum eigenvalue.
By Min-max theorem $$\lambda_n=\max_{x\in \mathbb{R^n},||{x}||=1   }x^TAx=\sum x_ix_j a_{ij}    $$
while on the other hand$$a_{ii}=e_i^TAe_i $$
 It follows that $$\lambda_n\ge a_{ii} \qquad \forall i$$
