The mean value theorem gives $\mathrm{\mid }\sqrt{x}-\sqrt{y}\mathrm{\mid }\le \frac{1}{\iota (2)}\mathrm{\mid }x-y\mathrm{\mid }$ for $x,y\ge 1$, so the square root is Lipschitz with constant $1\mathrm{/}2$ on $[1,\mathrm{\infty })$

Write $\sqrt{b}=b^{1\mathrm{/}2}$ for the nonnegative square root (Existence and uniqueness of $n$-th roots: a unique $a^{1\mathrm{/}n}\ge 0$ with $(a^{1\mathrm{/}n}{)}^n=a$, Rational powers $a^r$ of a positive base) and $\iota$ for the canonical natural (The canonical natural $\iota (n)=n\cdot 1_F$ of a field).

Claim. Let $I:=[1,\mathrm{\infty })$ and let $s:I\to \mathbb{R}$, $s(b):=\sqrt{b}$. Then

$\mathrm{\mid }\sqrt{x}-\sqrt{y}\mathrm{\mid }\le \frac{1}{\iota (2)}\mathrm{\mid }x-y\mathrm{\mid }\text{for all }x,y\in I,$

so $s$ is Lipschitz with constant $1\mathrm{/}\iota (2)$ on $I$ (Lipschitz map, $\alpha$-Hölder map for rational $0<\alpha \le 1$, and contraction, clause 3 of Dictionary: for $A\subseteq \mathbb{R}$ with the metric $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$, continuity and uniform continuity of $f:A\to \mathbb{R}$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $\mathbb{R}$ is compact in the open-cover sense of $\mathbb{R}$ exactly when it is a compact metric subspace) and hence uniformly continuous on $I$ (Uniform continuity of $f:A\to \mathbb{R}$: one $\delta$ serving every pair of points of $A$).

The constant is what the derivative bound gives, and the domain is what makes the bound available. On $[1,\mathrm{\infty })$ the derivative of $s$ is at most $1\mathrm{/}\iota (2)$; on $(0,1]$ it is not bounded at all, and the companion counterexample on this page shows that there the Lipschitz conclusion fails.

• Why the bound is $1\mathrm{/}2$ and not something smaller. The supremum of $s^{\mathrm{\prime }}$ over $(1,\mathrm{\infty })$ is approached at $b=1$, where $s^{\mathrm{\prime }}(1)=1\mathrm{/}\iota (2)$, and the argument uses nothing sharper than $b^{-1\mathrm{/}2}\le 1$. No claim is made that $1\mathrm{/}\iota (2)$ is the least Lipschitz constant on $I$; the corollary produces one constant that works, which is all the statement asserts.

• Everything depends on the left endpoint being $1$ and not $0$. The bound $b^{-1\mathrm{/}2}\le 1$ is exactly the statement $b\ge 1$, read through the monotonicity of rational powers. On $(0,1]$ the same expression is unbounded, and $x\mapsto \sqrt{x}$ on $(0,1]$ is differentiable with unbounded derivative and is not Lipschitz there, so the boundedness hypothesis in the Lipschitz corollary cannot be dropped on this page shows the conclusion then fails outright, so the hypothesis of If $f$ is continuous on an interval $I$ and $\mathrm{\mid }{f}^{\mathrm{\prime }}\mathrm{\mid }\le M$ at every interior point, then $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }\le M\mathrm{\mid }x-y\mathrm{\mid }$ for all $x,y\in I$, so $f$ is Lipschitz with constant $M$ and uniformly continuous on $I$ is doing real work here.

• The mean value theorem is inside the corollary, not applied directly. If $f$ is continuous on an interval $I$ and $\mathrm{\mid }{f}^{\mathrm{\prime }}\mathrm{\mid }\le M$ at every interior point, then $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }\le M\mathrm{\mid }x-y\mathrm{\mid }$ for all $x,y\in I$, so $f$ is Lipschitz with constant $M$ and uniformly continuous on $I$ is one application of The mean value theorem, as the case $g(x)=x$ of Cauchy's: for $f$ continuous on $[a,b]$ with $a<b$ and differentiable on $(a,b)$ there is $c\in (a,b)$ with $f(b)-f(a)=f^{\mathrm{\prime }}(c)(b-a)$ on the segment joining $x$ and $y$; quoting the packaged form avoids repeating the segment argument and, more importantly, avoids restating the endpoint conventions each time.