Prove that if two operators on a Hilbert space commute then the spectral radius of their sum is at most the sum of their spectral radii.
I tried to use Gelfand's formula along with the binomial formula to get somehow bound $\|(A+B)^n\|^{1/n}$ by $\|A^k\|^{1/k}+\|A^l\|^{1/l}$ where $l,k$ are functions of $n$ but have not been able to make progress yet.
The spectral radius remains unchanged when calculating it in the closed sub-algebra. So let's calculate the spectral radius in the unital abelian Banach algebra $\mathscr B$ generated by $A$ and $B$. \begin{align*}\sigma(A+B) & =\{\tau(A+B)|\tau \mbox{ is a character of } \mathscr B\}\\ & = \{\tau(A)+\tau(B)|\tau \mbox{ is a character of }\mathscr B\}\\ & \subset \sigma(A)+\sigma(B).\end{align*} It follows what you want.