Does the quantity
$$\sqrt{\frac{2n}{n + 1}}$$
have a global minimum, for large $n$?
Successive tries at WolframAlpha yield the following results:
Minimize $\sqrt{\frac{2n}{n + 1}}$ for $n > {10}^5$
Minimize $\sqrt{\frac{2n}{n + 1}}$ for $n > {10}^{10}$
Minimize $\sqrt{\frac{2n}{n + 1}}$ for $n > {10}^{15}$
Minimize $\sqrt{\frac{2n}{n + 1}}$ for $n > {10}^{20}$
Looking at the trend, it looks like the global minimum value for the quantity
$$\sqrt{\frac{2n}{n + 1}}$$
as $n$ becomes arbitrarily large is
$$\sqrt{2} \approx 1.414213562373.$$
Is there a way to prove this assertion without using limits?
As per request:
Notice that the function $f:\mathbb{R}_+\to\mathbb{R}_+$ given by the formula $\frac{2x}{x+1}$ is strictly increasing. For if $x,y\in\mathbb{R}_+$ with $x<y$, then $2x(y+1) = 2xy+2x < 2xy+2y = (x+1)2y$, and therefore, $f(x) = \frac{2x}{x+1} < \frac{2y}{y+1} = f(y)$.
We also note that the function $\sqrt{\cdot{}}:\mathbb{R}_+\to\mathbb{R}_+$ is strictly increasing (to prove it note that the derivative is $\frac{1}{2}x^{-\frac{1}{2}}$, which is strictly positive on $\mathbb{R}_+$. Furthermore, your function is $g(x)=\sqrt{\frac{2x}{x+1}}=\sqrt{f(x)}=(\sqrt{\cdot}\circ f)(x)$, which is thus a composition of strictly increasing functions and so is itself strictly increasing.
Hence, for all $x\in [b,\infty)$, $b\le x$, so $g(b)\le g(x)$. Therefore, $g$ has a global minimum on $[b,\infty)$, namely $g(b)$. We would like to know what happens as $b\to\infty$. But this just involves taking the limit:
$$ \lim_{b\to\infty} \min_{x\in[b,\infty)} g(x) = \lim_{b\to\infty} g(b) = \lim_{b\to\infty} \sqrt{\frac{2b}{b+1}} = \sqrt{2}. $$
However, this does not mean that $\sqrt{2}$ is a global minimum for $g(x)$ on some interval $[b,\infty)$. In fact, $\sqrt{2}$ is an upper bound on any such interval because $f(x)\le 2$, and so $g(x)=\sqrt{f(x)}\le\sqrt{2}$. It only means that $\liminf_{x\to\infty} g(x) = \sqrt{2}$.
The upshot of all this is that as posed, I don't think your question really makes any sense. I don't mean to cause offense, it is just my honest opinion.
Edit: I forgot to mention: on the interval $(b,\infty)$, $g$ has no minimum value since it is strictly increasing. More precisely, for every $x\in(b,\infty)$, there exists some $y\in(b,\infty)$ with $b<y<x<\infty$ so that $g(y)<g(x)$ since $g$ is strictly increasing. This shows that $g(x)$ is not a minimum value of $g$ on this interval. Furthermore, since $x\in (b,\infty)$ was arbitrary, this shows that $g$ has no minimum value on the interval $(b,\infty)$.