Determine whether the function $f(x) = \sqrt {x}$ is uniformly continuous on $A = [1, +\infty)$.

Question

Determine whether the function $f(x) = \sqrt {x}$ is uniformly continuous on $A = [1, +\infty)$.

The answer of the question is given below:

But I could not understand why $\sqrt{x} + \sqrt{y} > 2$, could anyone explain this for me please?

Answer 1

If $x$ and $y$ are both larger or equal to $1$, then $\sqrt{x} + \sqrt{y} \geq 1 + 1 = 2$.

Answer 2

Since $x,y\geqslant1$, you have $\sqrt x,\sqrt y\geqslant 1$ and therefore $\sqrt x+\sqrt y\geqslant2$.