How can I show $\limsup_n\frac{1}{n}\log(a_n+b_n)=\max\{\limsup_n\frac{1}{n}\log(a_n),\limsup_n\frac{1}{n}\log(b_n)\}$? I'm studing some arguments of dynamical system and the following property is used  a lot of time.
Let $a_n, b_n>0$ for every $n$ then 
$$\limsup_n\frac{1}{n}\log(a_n+b_n)=\max\{\limsup_n\frac{1}{n}\log(a_n),\limsup_n\frac{1}{n}\log(b_n)\}\\=\limsup_n\frac{1}{n}\log(\sqrt{a_n^2+b_n^2}).$$
How can I show that?
 A: Hint: For the first equality, we have that $\log(a_n+b_n)=\log\left(1+\frac{b_n}{a_n}\right)+\log(a_n)=$ $\log\left(1+\frac{a_n}{b_n}\right)+\log(b_n)$. Moreover, for each $n$, at least one of $\log\left(1+\frac{b_n}{a_n}\right)$ and $\log\left(1+\frac{a_n}{b_n}\right)$ is $\leq \log (2)$.
For the second equality, $\log(\sqrt{a_n^2+b_n^2})=\frac12 \log(a_n^2+b_n^2)$. Then use the first equality again and the property that $\log(x^2)=2\log(x)$.
P.S.: Notice that the $\limsup$'s don't change if we reset $a_n'=\min\{a_n,b_n\}$ and $b_n'=\max\{a_n,b_n\}$, which we still call $a_n$ and $b_n$. This might make proofs easier.
EDIT:

Proposition: Let $D\subset \mathbb{R}$, $f:D \longrightarrow \mathbb{R}$ be increasing and $(a_n),(b_n)$ be sequences in $D$. Then
$$\limsup_n f(\max\{a_n,b_n\})=\max\left\{\limsup_n f(a_n),\limsup_n f(b_n)\right\}$$

Proof: Since $f$ is increasing, $f(\max\{a_n,b_n\})=\max\{f(a_n),f(b_n)\}$. Now it suffices to use that $\limsup_n \max \{x_n,y_n\} = \max\{\limsup_n x_n, \limsup_n y_n\}$ (see, for instance, this answer). $\square$
