For $0≤a≤b≤c$, show that $\lim\sqrt[n]{a^n+b^n+c^n}=c$

Question

For $0≤a≤b≤c$, show that $\lim\sqrt[n]{a^n+b^n+c^n}=c$

I think I am making some silly mistake with my "proof". If it is indeed correct, another question emerges. So, my attempt is:

Since $0≤a≤b≤c$, it follows that $0≤a^n≤b^n≤c^n$. Thus, $a^n+b^n≤a^n+c^n$⇒$a^n+b^n+c^n≤a^n+2c^n≤3c^n$ which gives $\sqrt[n]{a^n+b^n+c^n}≤\sqrt[n]{3c^n}=c\sqrt[n]{3}$. So, the desired expression has an upper bound of $c\sqrt[n]{3}$.

We will seek to find a lower bound for it as well and use the "squeeze theorem".We seek a number $k\inℝ$ such that: $kc^n≤a^n+b^n+c^n$.We observe that for $k=\frac{1}2$ we get: $\frac12c^n≤a^n+b^n+c^n$⇒$c^n≤2a^n+2b^n+2c^n$⇒$0≤2a^n+2^n+c^n$ which is true for $0≤a≤b≤c$. It follows that $\sqrt[n]{\frac12c^n}≤\sqrt[n]{a^n+b^n+c^n}$

Thus we have a lower bound as well and:$\sqrt[n]{\frac12c^n}≤\sqrt[n]{a^n+b^n+c^n}≤c\sqrt[n]{3}$.

Since it holds that for every $t>0$ then $\lim\sqrt[n]{t}=1$, from the "squeeze theorem" we calculate the limits of the two bounds as: $\lim\sqrt[n]{\frac12c^n}=\lim c\sqrt[n]\frac12=\lim c\sqrt[n]{3}=c$ and we get the desired result.

Does this seem correct? And if so, what is the smallest $k$ we can use to get the lower bound and thus the result?

Answer 1

A faster way $c^{n} \leq a^{n}+b^{n}+c^{n} \leq c^{n}+c^{n}+c^{n}=3c^{n}$. Raising everything to the $\frac{1}{n}$ power, and applying squeeze, as well as $\lim 3^{\frac{1}{n}}=1$, gives the desired result.