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

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 $\lim_{n\to\infty} \sqrt[n]{m} = 1$.

Being $\{c_1,\cdots,c_m \}$ a fine set of number, exist and index $i\in [1,m]$ such that $$ \max\{c_1,\cdots,c_m \}=c_i. $$ Then results $$ \sqrt{c_1^n+\cdots+c_m^n} = \sqrt{c_i ^ n \Big(\frac{c_1^n}{c_i^n}+\frac{c_2^n}{c_i^n}+\frac{c_3^n}{c_i^n}+\cdots+1+\cdots+\frac{c_m^n}{c_i^n}\Big)} $$ where the number $1$ correpond to the $i$-th element. Being $c_i$ the max, then every ratio $c_1/c_i ,c_2/c_i,\cdots,c_m/c_i \leq 1$, i.e $$ \frac{c_j}{c_i}< 1,\,\forall j=1,\cdots,m,\, j\neq i $$ then for $n\to \infty$ follows $$ \Big(\frac{c_j}{c_i}\Big)^n\to 0,\, \forall j\neq i $$ So being $c_1>0$ then \begin{eqnarray} \lim_{n\to \infty}{\sqrt[1/n]{c_1^n+\cdots+c_m^n}}&=&\lim_{n\to\infty}{\sqrt{c_i ^ n \Big(\frac{c_1^n}{c_i^n}+\frac{c_2^n}{c_i^n}+\frac{c_3^n}{c_i^n}+\cdots+1+\cdots+\frac{c_m^n}{c_i^n}\Big)}}\\ &=&\lim_{n\to\infty}{c_i\sqrt{\frac{c_1^n}{c_i^n}+\frac{c_2^n}{c_i^n}+\frac{c_3^n}{c_i^n}+\cdots+1+\cdots+\frac{c_m^n}{c_i^n}}}\\ &=&\lim_{n\to\infty}{c_i\sqrt{\Big(\frac{c_1}{c_i}\Big)^n+\cdots+1+\cdots+\Big(\frac{c_m}{c_i}\Big)^n}}\\ &=&\lim_{n\to\infty}{c_i \cdot \sqrt{1}}\\ &=&c_i=\max\{c_1,\cdots,c_m\}. \end{eqnarray}

You can see there is an error in your approach if you consider a simple example. Let $c_1=2$ and $c_2=\cdots=c_m=0$. Then

$$\lim_{n\to\infty}\sqrt[n]{c_1^n+c_2^n+\cdots+c_m^n}=\lim_{n\to\infty}\sqrt[n]{2^n}=2$$ but

$$\lim_{n\to\infty}\sqrt[n]{\max\{c_1,c_2\ldots,c_m\}}=\lim_{n\to\infty}\sqrt[n]{2}=1$$

so your first inequality does not always hold.