Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$.
Let $A,B\in \mathcal{B}(F)$. Consider the following numbers: $$\alpha=\sup_{\substack{a,b\in \mathbb{C}^2,\\ |a|^2+|b|^2<1}}\sup\{|\mu|\,;\;\mu\in \sigma(aA+bB)\}.$$ $$\beta=\sup\{\lambda^{1/2}\,;\;\lambda\in \sigma(A^*A+B^*B)\},$$ where $\sigma(X)$ is the spectrum of an operator $X$.
I want to show that $$\alpha\geq \beta.$$
The inequality does not hold. Take $A$ any nonzero nilpotent element, and $B=0$. Then $\alpha=0$ and $\beta=\|A\|$.
The reverse inequality does hold, though. We have \begin{align} \operatorname{spr}(aA+bB)^2 &\leq \|aA+bB\|^2\leq (|a|\,\|A\|+|b|\,\|B\|)^2. \end{align} Then \begin{align} \sup_{a,b}\operatorname{spr}(aA+bB)^2 &\leq\sup_{a,b}(|a|\,\|A\|+|b|\,\|B\|)^2=\max\{\|A\|^2,\|B\|^2\}\\ &=\max\{\|A^*A\|,\|B^*B\|\}\\ &\leq \|A^*A+B^*B\|\\ &=\sup\{\lambda:\ \lambda\in\sigma(A^*A+B^*B)\}\\ &=\sup\{\lambda^{1/2}:\ \lambda\in\sigma(A^*A+B^*B)\}^2\\ \end{align} So $$ \alpha\leq\beta. $$