Bounds on spectral radius of $2\text{diag}(h)+h \cdot 1^T$ How can I get lower/upper bounds on the largest eigenvalue of the following sum of diagonal and rank-1 matrices for vector $h$ with $h_i>0\ \forall i$:
$$A=2\text{diag}(h)+h \cdot 1^T$$
For instance, for $d=3$ it would be matrix below
$$2 \left(
\begin{array}{ccc}
 h_1 & 0 & 0 \\
 0 & h_2 & 0 \\
 0 & 0 & h_3 \\
\end{array}
\right)+\left(
\begin{array}{ccc}
 h_1 & h_1 & h_1 \\
 h_2 & h_2 & h_2 \\
 h_3 & h_3 & h_3 \\
\end{array}
\right)
$$
The following has been observed to be an upper bound empirically
$$2\max_i h_i+\sum_i h_i\ge\lambda_\text{max}(A)$$
If we let $h=1,\frac{1}{2},\frac{1}{3},\ldots,\frac{1}{d}$, then for $d=4000$, the answer is $\approx 9.29455$, proposed upper bound is 10.8714. Furthermore, relative difference between bound and true value seems bounded as we vary $h$
Motivation: $\alpha<\lambda_1(A)$ is necessary and sufficient for the iteration $w=w-\alpha \langle w, x\rangle x$ to converge when $x$ is sampled from centered Normal with diagonal covariance and $h_i$ on the diagonal (derivation)

Some thoughts:
We deal with the case $h_1 > h_2 > \cdots > h_n$.
Consider the equation $Ax = \lambda x$ ($x\ne 0$) which is written as
$$(\lambda - 2h_k) x_k = h_k\sum_{i=1}^n x_i, \quad k=1, 2, \cdots, n. \tag{1}$$
We claim that $\lambda \ne 2h_j, \forall j$. Indeed,
if $\lambda = 2h_j$ for some $j$, then $\sum_{i=1}^n x_i = 0$ and $(2h_j - 2h_k)x_k = 0, \forall k\ne j$ which results in $x_k = 0, \forall k \ne j$. Then we get $x = 0$. Contradiction.
From $\lambda \ne 2h_j, \forall j$, we have $\sum_{i=1}^n x_i \ne 0$.
From (1), we have
$$x_k = \frac{h_k}{\lambda - 2h_k}\sum_{i=1}^n x_i, \quad k=1, 2, \cdots, n. $$
Thus, we have
$$\sum_{k=1}^n \frac{h_k}{\lambda - 2h_k} = 1. \tag{2}$$
Fact 1: The equation (2) has exactly $n$ distinct real solutions $\lambda_1 > \lambda_2 > \cdots > \lambda_n$
with $\lambda_1 > 2h_1$ and $\lambda_k \in (2h_k, 2h_{k-1}), k=2, 3, \cdots, n$.
(The proof is easy and thus omitted here.)
Let us give a lower bound of $\lambda_1$.
We have
$$\frac{h_k}{\lambda_1 - 2h_k} = \frac{h_k}{\lambda_1}\cdot \frac{1}{1 - 2h_k/\lambda_1} > \frac{h_k}{\lambda_1} \cdot \left(1 + \frac{2h_k}{\lambda_1}\right), \quad \forall k.$$
Thus, we have
$$\frac{\sum_{i=1}^n h_i}{\lambda_1} + \frac{2\sum_{i=1}^n h_i^2}{\lambda_1^2} < 1$$
which results in
$$\lambda_1 > \frac12 \sum_{i=1}^n h_i
+ \frac12\sqrt{\left(\sum_{i=1}^n h_i\right)^2 + 8 \sum_{i=1}^n h_i^2}. \tag{3}$$
When $h = 1, \frac12, \frac13, \cdots, \frac1d$ and $d = 4000$, (3) gives $\lambda_1 > 9.227851206$. Using Maple, from (2), we get $\lambda_1 \approx 9.294554415$.
A better lower bound:
We have
$$\frac{h_1}{\lambda_1 - 2h_1} + \frac{\sum_{i=2}^n h_i}{\lambda_1} + \frac{2\sum_{i=2}^n h_i^2}{\lambda_1^2} < 1. \tag{4}$$
When $h = 1, \frac12, \frac13, \cdots, \frac1d$ and $d = 4000$, (4) gives $\lambda_1 > 9.284803103$.
