Largest value of $\lambda_1$ in $\frac{h_1}{\lambda_1 - 2h_1} + \frac{\sum_{i=2}^n h_i}{\lambda_1} + \frac{2\sum_{i=2}^n h_i^2}{\lambda_1^2} \le 1.$ Suppose $h$ is a probability distribution over $d$ outcomes, and $\lambda(h)$ is the largest value of $\lambda_1$ satisfying the following inequality:
$$\frac{h_1}{\lambda_1 - 2h_1} + \frac{\sum_{i=2}^n h_i}{\lambda_1} + \frac{2\sum_{i=2}^n h_i^2}{\lambda_1^2} \ge 1. $$
Can we say anything about the value of $\lambda(h)$ when effective rank $R(h)=\frac{(\sum h_i)^2}{\sum h_i^2}$ is large?
Empirically, this value seems to approach $\|h\|_1 + 2 \|h\|_\infty$ as we increase $R$

This is a follow-up on earlier related question . Proving this would show that matrix $l_1$-norm and spectral radius coincide for high-dimensional diagonal+rank1 matrix $A=2\text{diag}(h)+h \otimes 1^T$
 A: The conjecture that the value approaches $\|h\|_1 + 2\|h\|_\infty$ is false. I show below that for every $h$ such that $\|h\|_\infty \le 1/4,$ the maximum value is at most $\|h\|_1 + \frac{3}{2} \|h\|_\infty + \frac{1}{2R}.$

Let me use $\mathbf{h}$ to denote the probability vector, and just $h$ to denote $\mathbf h_1 = \|\mathbf h\|_\infty$ (where I'm using that in the linked answer, $h_1 \ge h_2 \ge \cdots$). Similarly, let me just write $\lambda$ instead of $\lambda_1$.
Since $\mathbf{h}$ is a probability vector, $\sum \mathbf{h}_i = 1$, and $R(\mathbf{h}) = 1/\sum \mathbf{h}_i^2$. Using this,  we can simplify the  inequality to $$ \varphi(\lambda;h,\rho) := \frac{h}{\lambda - 2h} + \frac{1-h}{\lambda} + \frac{2 (\rho - h^2)}{\lambda^2} \ge 1,$$ where $\rho = R^{-1} \in [h^2, h]$ is small. It's worth being careful about what small means here: for instance, if $\rho \ll h,$ as can happen when $h = 1/\sqrt{d}, \rho \approx 2/d,$ it probably doesn't make sense to say that $h$ is small (even though it is $o(1)$ as $d$ grows) since the behaviour we are interested in is sensitive at least to the first order in $h$. So I'll take small to be $o(h)$ (and of course, large $R$ means $\omega(1/h)$)
Let $\lambda(h,\rho)$ be the maximal value where $\varphi \ge 1$ holds. It's easy to see that $\lambda(h,\rho) \le 1 + 2h,$ and that if $h > 2\rho$ that $\lambda(h,\rho) \ge 1.$ The conjecture then is that in the regime $\rho = o(h),$ $\lambda(h,\rho) = 1 + 2h - o(h)$.
Since cubics are relatively easy to manipulate with a CAS, to bring further behaviour out, I'm going to multiply through by $\lambda^2(\lambda-2h),$ and rewrite the condition as finding the largest $\lambda > 2h$ such that $ f(\lambda;h,\rho) \le 0, $ where $$ f(\lambda;h,\rho) := \lambda^3 - (1+2h)\lambda^2 + 2(h-\rho)\lambda + 4h(\rho -h^2).$$ Note that $f$ has a unique root in $(2h, \infty)$, and this root is the value of $\lambda(h,\rho)$ (The cleanest way to see this is by observing that $1-\varphi$ has a unique root in $(2h,\infty),$ due to the monotonicity of $\varphi$ and its behaviour near $2h$, and of course, $\lambda^2 (\lambda - 2h)$ has no roots in this region, so $f = \lambda^2(\lambda-2h)(1-\varphi)$ inherits the roots of $1-\varphi$ in $(2h,\infty)$).
It's a direct computation (see this) that $$ f(1 + (3/2)h + (1/2)\rho; h,\rho) \\ = \frac{3}{2}(h-\rho) + \frac{3}{2}h^2+ 3h\rho - \frac{\rho^2}{2} - \frac{41}{8} h^3 + \frac{3}{8}h^2\rho + \frac{5}{8}h\rho^2 + \frac{\rho^3}{8} $$ Now, if $h < 1/4,$ then observe that $3/2 h^2 > 41/8 h^3,$ and of course $3h\rho \ge 3\rho^2 \ge \rho^2/2.$ Thus, for $h < 1/4,$ it holds that $f(1 + (3/2)h + (1/2) \rho) > 0 \implies \lambda(h,\rho) \le 1 + 3/2 h + \rho/2.$ Note that this is independent of the relative scaling of $h, \rho$, so a fortiori this also holds for every $h < 1/4$ even if $\rho \ll h$.
I suspect that the behaviour of $\lambda(h, \rho)$ can be fairly rich to the first order. For instance, a similar direct computation shows that if $\rho > 2h^2,$ then $f(1+h + \rho) > 0,$ and if $\rho \gg h^{3/2},$ then $\lambda(h,\rho) \le 1 + 2\rho$. As for lower bounds, it does follow from the same trick that for every $h < 1/2, \lambda(h,\rho) > 1 + \rho. $
