Question

Why is the following true?\[|\sqrt{x}-\sqrt{y}| \le \sqrt{|x-y|}\]

Answer 1

one way seem sot be a proof by exhaustion; try all four cases...

Answer 2

seems to*

Answer 3

Triangle inequality kind of thing?

Answer 4

\[\sqrt x-\sqrt y\le\sqrt{y-x}\]\[x-2\sqrt{xy}+y\le y-x\]\[-2\sqrt{xy}\le-2x\]\[\sqrt{xy}\ge x\implies\frac yx\ge1\implies y\ge x\]or\[\sqrt x-\sqrt y\le\sqrt{x-y}\]\[x-2\sqrt{xy}+y\le x-y\]\[-2\sqrt{xy}\le-2y\]\[\sqrt{xy}\ge y\implies\frac xy\ge1\implies x\ge y\]this means that either \(x=y=1\) or that either \(x>y\vee x< y\)
the same argument follows for the other two cases

Answer 5

I didn't really employ the triangle inequality, but I bet you could
I just did it the way that seemed most obvious to me; looking for a contradiction

Answer 6

okay my above doesn't mean that x=y=1, just that x=y or x>y or y>x

Answer 7

still I think it is convincing, but I suppose you are unsatisfied...

Answer 8

Looks like Holder's inequality