Uniform convergence of a n-th root on $\mathbb{R}^+$

This is closely related to Study the pointwise convergence and uniform convergence of the sequence of functions $f_n(x)=\sqrt[n]{1+x^{2n}}$.

Since $\max(1,u)\le\left(1+u^n\right)^{1/n}\le\left(n^2+u^n\right)^{1/n}$, uniform convergence in this question implies uniform convergence in that question.

I wrote the following answer soon after this question was asked, but I was waiting for a response from the author to comments. However, since the linked question has been asked, it seems like a good time to post this answer

Assume that $u\gt0$ and $n\ge2$.

For $u^n\le n^2$, $$ \begin{align} \left(1+\tfrac1{\sqrt{n}}\right)^n&\ge1+\sqrt{n}\tag{1a}\\ \left(1+\tfrac1{\sqrt{n}}\right)^{4n} &\ge\left(1+\sqrt{n}\right)^4\tag{1b}\\[6pt] &\ge n^2\tag{1c}\\[6pt] \left(1+\tfrac1{\sqrt{n}}\right)^4&\ge\left(n^2\right)^{1/n}\tag{1d}\\[3pt] 2^{1/n}\left(1+\tfrac1{\sqrt{n}}\right)^4&\ge\left(n^2+n^2\right)^{1/n}\tag{1e}\\[4pt] 2^{1/n}\left(1+\tfrac1{\sqrt{n}}\right)^4&\ge\left(n^2+u^n\right)^{1/n}\tag{1f} \end{align} $$ Explanation:
$\text{(1a)}$: Bernoulli's Inequality
$\text{(1b)}$: raise to the fourth power
$\text{(1c)}$: $\left(1+\sqrt{n}\right)^4\ge\left(\sqrt{n}\right)^4=n^2$
$\text{(1d)}$: take the $n^\text{th}$ root
$\text{(1e)}$: multiply by $2^{1/n}$
$\text{(1f)}$: $u^n\le n^2$

For $u^n\ge n^2$, $$ \begin{align} \left(u^n+n^2\right)^{1/n} &=u\left(1+\frac{n^2}{u^n}\right)^{1/n}\tag{2a}\\ &\le u\left(1+\frac{n}{u^n}\right)\tag{2b}\\[3pt] &\le u+u^{1-n/2}\tag{2c}\\[9pt] &\le u+n^{2/n-1}\tag{2d} \end{align} $$ Explanation:
$\text{(2a)}$: distributive property
$\text{(2b)}$: Bernoulli's Inequality
$\text{(2c)}$: $n\le u^{n/2}$
$\text{(2d)}$: $u\ge n^{2/n}$ and $1-n/2\le0$

Choose any $\epsilon\gt0$. Since both $2^{1/n}\left(1+\tfrac1{\sqrt{n}}\right)^4-1$ and $n^{2/n-1}$ tend to $0$ as $n\to\infty$, we can find an $N$ so that for $n\ge N$, $$ \max\left(2^{1/n}\left(1+\tfrac1{\sqrt{n}}\right)^4-1\ ,\ n^{2/n-1}\right)\le\epsilon\tag3 $$ then $(1)$, $(2)$, and $(3)$ guarantee that for $n\ge N$, $$ \max(1,u)\le\left(n^2+u^n\right)^{1/n}\le\max(1,u)+\epsilon\tag4 $$ Thus, setting $u=x^4$, we get that $$ \lim_{n\to\infty}\sup_{x\ge0}\left|\left(n^2+x^{4n}\right)^{1/n}-\max\left(1,x^4\right)\right|=0\tag5 $$ That is, the convergence is uniform.