Prove that $\lim \limits_{n \to \infty}{\sqrt[n]{a^n+b^n}}=\max(a,b)$ if $(a_n,b_n)\to(a,b)$

Question

I know well that:

$$ \max(x,y)=\frac{x+y+\lvert y-x\rvert}{2}$$

but I do not see how that it would be useful.

Answer 1

Without loss of generality, let $a \geq b$. Then

$$ a \leq \sqrt[n]{a^n + b^n} \leq \sqrt[n]{a^n + a^n} = 2^{1/n} a$$

Now apply the sandwich theorem.

Answer 2

Hint: For positive $a$ and $b$, note that $$(a^n+b^n)^{1/n} = a\left(1+\left(\tfrac ba\right)^n\right)^{1/n}$$ and $$(a^n+b^n)^{1/n} = b\left(1+\left(\tfrac ab\right)^n\right)^{1/n}$$