Convergence of $\sqrt[n]{x^n+y^n}$ (for $x, y > 0$)

How can I prove the convergence of the sequence $b_n=\sqrt[n]{x^n+y^n}$ where $x, y > 0$?

Can I divide it in two cases?

Case 1: $x > y$.

$$ b_n=\sqrt[n]{x^n+y^n} < \sqrt[n]{x^n+x^n} = \sqrt[n]{2 \cdot x^n}=x \cdot \sqrt[n]{2} $$

Case 2: $x < y$.

$$ b_n=\sqrt[n]{x^n+y^n} < \sqrt[n]{y^n+y^n} = \sqrt[n]{2 \cdot y^n}=y \cdot \sqrt[n]{2}$$

Result: Does the sequence converges to $\max(x,y)$?

Your suggestion of breaking up into cases is the right one. I would handle the details differently. Suppose for example that $x<y$. We use the fact that $$\sqrt[n]{x^n+y^n}=y\sqrt[n]{1+\left(\frac{x}{y}\right)^n}.$$

Since $\dfrac{x}{y}<1$, it is clear that $\displaystyle\lim_{n\to\infty} \left(\frac{x}{y}\right)^n=0$.

So if $x<y$, the limit is $y$. Similarly, if $y<x$, the limit is $x$. We need also to take care of the case $y=x$. Then we want to calculate $\displaystyle\lim_{n\to\infty} x\sqrt[n]{2}$. As $n$ gets large, $\sqrt[n]{2}$ approaches $1$.

The answer is therefore always, as you conjectured, $\max(x,y)$.

Comment: We can also handle the problem without breaking into cases. Note that $$\max(x,y) <\sqrt[n]{x^n+y^n} \le \sqrt[n]{2}\max(x,y).$$ Then we can obtain the desired limit formula directly by "squeezing," since $\displaystyle\lim_{n\to\infty}\sqrt[n]{2}=1$. This argument is more elegant but perhaps less natural than the division procedure that was used in the main post. The intuition there was that if $x<y$, then for large $n$, the term $x^n$ is negligible compared to $y^n$.

Just to complement what has been said, note that these arguments typically yield the equivalence of the $L^n$ norms and $sup$ norm for a finite dimensional vector space.

Namely, for $\boldsymbol{x}=(x_1,\ldots,x_d)\in\mathbb{R}^d$, $d\geq 1$, it is usual to consider the norms $\|\boldsymbol{x}\|_n=(\sum_{i=1}^d|x_i|^n)^{1/n}$ for any $n\geq 1$ and $\|\boldsymbol{x}\|_{\infty}=\sup_{i=1}^d|x_i|$.

Then for all $n\geq 1$, $\|\boldsymbol{x}\|_\infty\leq \|\boldsymbol{x}\|_n \leq d^{1/n}\|\boldsymbol{x}\|_n$.

This moreover gives for any $d\geq 1$

$$ \lim_{n\rightarrow\infty}(\sum_{i=1}^d|x_i|^n)^{1/n} = \lim_{n\rightarrow\infty}\|\boldsymbol{x}\|_n = \|\boldsymbol x\|_\infty=\max_{i=1}^d|x_i|, $$ your case being $d=2$.