Relationship between Difference of Two Numbers and Their Square Roots

Question

Is there a relationship between the difference of two numbers and the difference of their square roots? For example, can we say that

${| \sqrt x - \sqrt y|\leq |x - y|}$ when ${ x, y \geq 1 }$,

but

${| \sqrt x - \sqrt y|\geq |x - y|}$ when ${ x, y \leq 1 }$?

Answer 1

Unless $x$ and $y$ are both $0$, we have $$|\sqrt{x}-\sqrt{y}|=\frac{|x-y|}{\sqrt{x}+\sqrt{y}}.$$ It follows that if $\sqrt{x}+\sqrt{y}\ge 1$ then we have $|\sqrt{x}-\sqrt{y}|\le |x-y|$, and if $\sqrt{x}+\sqrt{y}\le 1$ then we have $|\sqrt{x}-\sqrt{y}|\ge |x-y|$.

So the right condition is not quite the one in the post. If either $x\ge 1$ or $y\ge 1$, then the first inequality holds. But it also holds for certain $x$ and $y$ both below $1$, for example if both $x$ and $y$ are $\ge \frac{1}{4}$.

Answer 2

If $x,y>0$ then $$(\sqrt x-\sqrt y)(\sqrt x+\sqrt y)=x-y$$ and so $$|\sqrt x-\sqrt y|=\frac{|x-y|}{|\sqrt x+\sqrt y|}\ .$$ So $|\sqrt x-\sqrt y|\le|x-y|$ if $\sqrt x+\sqrt y\ge1$, and the reverse if $\sqrt x+\sqrt y\le1$.

Answer 3

$x$ and $n$ are real numbers, and $n$ is the distance between $x$ and (duh) ($x-n$).

$x^2-(x-n)^2=2nx-n^2$

Example: $26^2-22^2$ (without a calculator and you don't want to square $26$ and $22$)$= 2(4)(26)-4^2=192$

P.S. I'm in tenth grade, and I spent an hour figuring this out myself. This works especially well for very large numbers whose square roots are gross, and you don't have a calculator.