Let $T_1$ and $T_2$ be two bounded linear operators in a complex banach space $X$.
If $T_1T_2 = T_2T_1$, I want to know how to show that $$ r(T_1+T_2) \leq r(T_1) + r(T_2), $$ where $r(A)$ denotes the spectral radius of $A$, and it is given by $$ r(A) = \inf_{n>0} \|A^n\|^{1/n} = \lim_{n\rightarrow \infty} \|A^n\|^{1/n}. $$
For simplicity, assume that $\|T\| \leqslant 1$ and $\|S\| \leqslant 1$. Write $$ (T + S)^n = \sum_{k = 0}^n \tbinom{n}{k} T^k S^{n - k} . $$ Fix $\varepsilon > 0$, and let $m$ be large enough, so that $\|T^j\| \le (r(T) + \varepsilon)^j$ and $\|S^j\| \le (r(S) + \varepsilon)^j$ for $j \ge m$. Then, if $n > 2 m$, we have $$ \begin{aligned} \|(T + S)^n\| & \leqslant m \tbinom{n}{m} (\|T^{n - m}\| + \|S^{n - m}\|) + \sum_{k = m}^{n - m} \tbinom{n}{k} (r(T) + \varepsilon)^k (r(S) + \varepsilon)^{n - k} \\ &\leqslant m \tbinom{n}{m} ((r(T) + \varepsilon)^{n - m} + (r(S) + \varepsilon)^{n - m}) + (r(T) + r(S) + 2 \varepsilon)^n . \end{aligned} $$ Since $(m \tbinom{n}{m})^{1/n} \to 1$ as $n \to \infty$, we easily find that $$ r(T + S) \leqslant \max\{r(T) + \varepsilon, r(S) + \varepsilon, r(T) + r(S) + 2 \varepsilon\} = r(T) + r(S) + 2 \varepsilon . $$ This gives the desired inequality.
