Change of the smallest positive eigenvalue after a rank-one update

Question

Given natural numbers $n,r,R\in\mathbb{N}$ with $r,R\le n$, let $A\in\mathbb{R}^{n\times r}$ and $B\in \mathbb{R}^{n\times R}$ be two matrices with full column rank and let $c\in\mathbb{R}^n$. Denote by $\lambda_{\mathrm{min}^\star}(X)$ the smallest positive eigenvalue of $X\in\mathbb{R}^{n\times n}$. Is it true that $$ \lambda_{\mathrm{min}^\star}(AA^\intercal + BB^\intercal + cc^\intercal) - \lambda_{\mathrm{min}^\star}(AA^\intercal + BB^\intercal) \le \lambda_{\mathrm{min}^\star}(AA^\intercal + cc^\intercal) - \lambda_{\mathrm{min}^\star}(AA^\intercal) ? $$ If it is not true, are there conditions on $A$, $B$ and $c$ such that it is?

Answer 1

It is not true, try $$A=\left( \begin{array}{cc} 2 & 0 \\ 0 & 1 \\ \end{array} \right),\;\;B=\left( \begin{array}{cc} 1 & 0 \\ 0 & 3 \\ \end{array} \right),\;\;c=\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right),$$ $$\lambda_{\mathrm{min}^\star}(AA^\intercal + BB^\intercal + cc^\intercal) - \lambda_{\mathrm{min}^\star}(AA^\intercal + BB^\intercal)=6-5=1,$$ $$\lambda_{\mathrm{min}^\star}(AA^\intercal + cc^\intercal) - \lambda_{\mathrm{min}^\star}(AA^\intercal)=1-1=0.$$