So there is a proof of the inequality needed:
$|\sqrt{x}-\sqrt{y}| \leq \sqrt{|x-y|} $ , where $x$,$y$ $\geq$ $0$
After squaring both sides:
$(|\sqrt{x}-\sqrt{y}|)^2 \leq (\sqrt{|x-y|})^2$
$x + y – 2\sqrt{xy} \leq |x-y|$,
Apparently, now you need cases of:
Case 1: $x \geq y$
Case 2: $x > y$
Case 3: $y > x$
I am asking why you need cases (in general for a proof and specifically this one) and also how would case 3 work? Thanks.
Best Answer
We can consider two main cases, indeed we need $x,y \geq 0$ and
$$|\sqrt{x}-\sqrt{y}| \leq \sqrt{|x-y|} \iff \sqrt{x}-\sqrt{y} \leq \sqrt{x-y}$$
$$|\sqrt{x}-\sqrt{y}| \leq \sqrt{|x-y|} \iff -\sqrt{x}+\sqrt{y} \leq \sqrt{-x+y}$$
and since in both cases LHS and RHS are positive we can proceed by squaring both sides.