real analysis - if $0

if $0

$\mid \sqrt[n]{a^n+b^n+c^n} - c \mid \, < \,\mid \sqrt[n]{a^n}+\sqrt[n]{b^n}+\sqrt[n]{c^n} - c \mid$ doesn't get me anywhere.

$$\begin{align}
\ln((a^n+b^n+c^n)^{1/n}) &= \frac{1}{n}\ln(a^n+b^n+c^n) \\
&= \frac{1}{n}\ln\left(c^n\left(\frac{a^n}{c^n} + \frac{b^n}{c^n} + 1\right) \right)\\
&= \frac{1}{n}\bigg (\ln(c^n)+\ln(\frac{a^n}{c^n} + \frac{b^n}{c^n} + 1)\bigg) \\
&\le \frac{1}{n}\bigg( \ln(c^n) + \ln(3) \bigg) \\&= \frac{1}{n}\bigg( n \cdot \ln(c) + \ln(3) \bigg)
\end{align}$$

$\mid \ln(\sqrt[n]{a^n+b^n+c^n} - \ln(c)\mid \,\le\, \mid\frac{1}{n} \mid \mid (n\cdot \ln(c) + \ln(3)) - n\cdot \ln(c)\mid = \mid ( ln(c) + \frac{ln(3)}{n}) - \ln(c)\mid $

let $\epsilon > 0$
choose $N \in \mathbb{N}$ such that $N > \frac{1}{\epsilon}$ then $\forall n > N \implies \mid ( \ln(c) + \frac{ln(3)}{n}) - \ln(c)\mid < \epsilon$