real analysis - Show that $lim_{nrightarrow infty} sqrt[n]{c_1^n+c_2^n+ldots+c_m^n} = max{c_1,c_2,ldots,c_m}$

Let $m\in \mathbb{N}$ and $c_1,c_2,\ldots,c_m \in \mathbb{R}_+$. Show that

$$\lim_{n\rightarrow \infty} \sqrt[n]{c_1^n+c_2^n+\ldots+c_m^n} = \max\{c_1,c_2,\ldots,c_m\}$$

My attempt: Since

$$\lim_{n\rightarrow \infty} \sqrt[n]{c_1^n+c_2^n+\ldots+c_m^n} \leq \lim_{n\rightarrow \infty}\sqrt[n]{\max\{c_1,c_2,\ldots,c_m\}} = \lim_{n\rightarrow \infty} \sqrt[n]{n}\sqrt[n]{\max\{c_1,c_2,\ldots,c_m\}}=\lim_{n \rightarrow \infty} \max\{\sqrt[n]{c_1^n},\sqrt[n]{c_2^n},\ldots,\sqrt[n]{c_m^n}\}=\lim_{n \rightarrow \infty}\max\{c_1,c_2,\ldots,c_m\}=\max\{c_1,c_2,\ldots,c_m\}$$

it follows that $\lim_{n\rightarrow \infty} \sqrt[n]{c_1^n+c_2^n+\ldots+c_m^n}$ is bounded, but I don't think it's monotonically decreasing, at least I can't prove this. Can anybody tell me whether the approach I have chosen is a good one, whether what I have done is correct and how to finish the proof?

Answer

The short proof.

Let $c=\max\{c_1,c_2,\dots,c_m\}$ and note that:

$$c^n \leq c_1^n+c_2^n+\dots+c_m^n \leq mc^n$$

Now take the $n$th root, and see that $\lim_{n\to\infty} \sqrt[n]{m} = 1$.