I am given a different definition of a finite order of an entire function from the one given in the link Order of $\frac{f}{g}$. My definition is that an entire function $f$ has finite order $\rho$ if $$\rho = \inf \{a : |f(z)| \le \exp(|z|^a) \quad \forall |z| \ge r, \text{ some } r > 0 \}.$$ This is equivalent to saying, letting $\epsilon > 0$, we have $|f(z)|< \exp(|z|^{\rho + \epsilon})$ for all $z$ with $|z|$ sufficiently large.
I would like to use these to show that if $\rho_1$, $\rho_2$, $\rho$ are of finite orders of entire functions $f_1$, $f_2$, and $f_1 + f_2$ respectively, then $\rho \le \max(\rho_1, \rho_2)$.
Let $M(r) = \max_{|z|=r} |f(z)|$ and $\rho_0 = \max(\rho_1, \rho_2)$. Then for $r = |z|$ sufficiently large, $$M(r) < \exp(|z|^{\rho_1 + \epsilon}) + \exp(|z|^{\rho_2 + \epsilon}) \le 2 \exp(|z|^{\rho_0 + \epsilon}).$$ I'm getting stuck at this step. What should I do next? I'm thinking of writing $2 = e^{\log 2}$.
Thank you.
$$ 2 \exp(|z|^{\rho_0 + \epsilon})=2\exp(|z|^{\rho_0 + \epsilon}(1-|z|^\epsilon)) \exp(|z|^{\rho_0 + 2\epsilon})\le\exp(|z|^{\rho_0 + 2\epsilon}) $$ if $|z|$ is large enough. In fact, it is enough to have $$ |z|^{\rho_0 + \epsilon}(1-|z|^\epsilon)\le\log\frac12, $$ and this is possible because $|z|^{\rho_0 + \epsilon}(1-|z|^\epsilon)\to-\infty$ as $|z|\to\infty$.