The Spectral Radius Inequality: $r(A+B)\leq r(A)+r(B)$

Let $A,B$ be two elements in a Banach Algebra, the spectral radius is defined by $$r(A):=\limn \sqrt[n]{\|A^n\|},$$ and it is a standard fact that the limit above exists (therefore $r(A)$ is properly defined). Recently (in office) I am asked to try:

Problem. If $AB=BA$, show that $r(A+B)\leq r(A)+r(B)$.

This problem is stated in a problem book and described as an easy consequence of the definition of $r(A)$ without proof, we work out the detail:

Proof. 
Since $r(cA)=cr(A)$ for any $c\ge 0$, we can assume that $\|A\|,\|B\|<1$. Now there is an $N$ such that for every $k,n$ such that $k\ge N$ and $n-k\ge N$,
$$\|A^{n-k}\|\leq (r(A)+\epsilon)^{n-k}\quad \text{and}\quad \|B^k\|\leq (r(B)+\epsilon)^k.$$ Here $\epsilon$ can be arbitrarily small such that both $r(A)+\epsilon,r(B)+\epsilon<1$.

Now we multiply them and take summation to get
$$\bigg\|\sum_{k\ge N\atop n-k\ge N} \binom{n}{k}A^{n-k}B^k\bigg\|\leq \sum_{k=N}^{n-N}\binom{n}{k}(r(A)+\epsilon)^{n-k}(r(B)+\epsilon)^k$$ and if we add both sides by what's missing on the left in $(A+B)^n$ and also add the remaining terms of the series on the RHS, we have
$$\|(A+B)^n\|\leq \bigg\|\sum_{k>n-N,\text{ or}\atop k<N}\binom{n}{k}A^{n-k}B^k\bigg\|+ (r(A)+r(B)+2\epsilon)^n.$$ Now let's split the summation into two parts, the RHS above
$$\begin{align*} &\leq \begin{array}{c}\displaystyle  \sum_{k>n-N} \binom{n}{k}\|A^{n-k}\|\|B^k\| +\sum_{k<N}\binom{n}{k}\|A^{n-k}\|\|B^k\| \\ +(r(A)+r(B)+2\epsilon)^n \end{array}\\ &\leq \begin{array}{c} \displaystyle  \sum_{k>n-N}\binom{n}{k}\cdot 1 \cdot (r(B)+\epsilon)^k+\sum_{k<N}\binom{n}{k}\cdot (r(A)+\epsilon)^{n-k}\cdot 1\\  +(r(A)+r(B)+2\epsilon)^n \end{array}\tag*{$(*)$} \\ &\leq N\binom{n}{N}(r(B)+\epsilon)^{n-N}+ N\binom{n}{N}(r(A)+\epsilon)^{n-N}+(r(A)+r(B)+2\epsilon)^n\\ &\leq 2N\binom{n}{N} (r(A)+r(B)+2\epsilon)^{n-N}+(r(A)+r(B)+2\epsilon)^n\\ &=\brac{2N\binom{n}{N}(r(A)+r(B)+2\epsilon)^{-N}+1} (r(A)+r(B)+2\epsilon)^n, \end{align*}$$ we assume that in $(*)$ $n$ is is big enough to make sure $n-N>N$. Now we take $n$th root and take $\lim$ on both sides ($\binom{n}{N}$ is just a scalar times finite many linear factors in $n$) to get
$$r(A+B)\leq r(A)+r(B)+2\epsilon.$$
Finally set $\epsilon\to 0^+$.$\qed$