Min or max

Without loss of generality, say the maximum of $a,b,c$ is $a$ . Then $\left(a^p+b^p+c^p \right)^{\frac{1}{p}}=a\left(1+\left(\frac{b}{a}\right)^p+ +\left(\frac{c}{a}\right)^p \right)^{\frac{1}{p}}$

Now, the (positive) quantity $\varepsilon=\left(\frac{b}{a}\right)^p+ \left(\frac{c}{a}\right)^p$

can be made arbitrarily small by choosing $p$ large enough.

So $\left(a^p+b^p+c^p \right)^{\frac{1}{p}}=a(1+\varepsilon)^{\frac{1}{p}}=a+\frac{1}{p}a\varepsilon+\mathcal{O} \left( \varepsilon^2 \right) \to a$

ie the quantity in the question tends to $\max(a,b,c)$ as $p \to \infty$ .

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Incidentally, this is a very useful identity and very common in analysis of dynamical systems (where you are often interested in long term behaviour of a differential equation near a fixed point).

For similar reasons, this is also why the $n^{\text{th}}$ Fibonacci number is well approximated by $\varphi^n$ .

We shall in general show that, for $\lambda_i\in\mathbb R^n$ , $i=1,2,3,\cdots, n$ $\operatorname{max}\left\{|\lambda_1|,\lambda_2,\cdots,|\lambda_n| \right\}=\lim_{p\to\infty}\left(\sum_{i=1}^n|\lambda_i|^p\right)^{1/p}$

Proof

Let $\mathcal{A}_i=\left\{|\lambda_1|,|\lambda_1|,\cdots ,|\lambda_n|\right\}$ be a finite set such that for $i\in [1,n]$ , $\operatorname{max}\left\{|\lambda_1|,|\lambda_1|,\cdots ,|\lambda_n|\right\}= |\lambda_k|$ for any $1\leq k\leq i\leq n$ then we can write $\mathcal{S}(n,k)=\sum_{i=1}^{n}|\lambda_i|^p=|\lambda_k|^p\sum_{i=1}^n\Bigg|\frac{\lambda_i}{\lambda_k}\Bigg|^p=|\lambda_k|^p\left(1+\sum_{i=1,i\neq k}^n\Bigg|\frac{\lambda_i}{\lambda_k}\Bigg|^p\right)$ now with pth root of $\mathcal{S}(n,k)$ and by Bernoulli inequality we have for $0<j<1$ $|\lambda_k|\leq \sqrt[p]{\mathcal{S}(n,k)}\leq |\lambda_k|\left(1+\frac{1}{p}\sum_{i=1,i\neq k}^n\Bigg|\frac{\lambda_i}{\lambda_k}\Bigg|^p\right)$ and hence as $p\to \infty$ , $\left(\frac{\lambda_i}{\lambda_k}\right)^p=j^p\to 0$ as $j<1$ and by Sandwich theorem we have $|\lambda_k|\leq \lim_{p\to\infty}\sqrt[p]{\mathcal{S}(n,k)}\leq |\lambda_k|$ and hence $\operatorname{max}\left\{A_i\right\}=\lim_{p\to \infty+}\left(\sum_{i=1}^n\mathcal{A}_i^p\right)^{1/p}$

For our case we just set $|\lambda_1|=a ,|\lambda_2| =b$ and $|\lambda_3|=c$ and hence required answer is $\operatorname{max}(a,b,c)$ .

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Remark: It is interesting to note that for all $x,y\in\mathbb R^n$ $d_{\infty}(x,y)=\lim_{p\to \infty}d_p(x,y)$ where $d_{\infty}(x,y)$ is called sup-norm metric and $d_p(x,y)=\left(|x_1-y_1|^p+|x_2-y_2|^p\right)^{1/p}$ which is directly a implication of above result.