Proofs using absolute value with Triangle/AGM Inequality

[darksteel88]

Homework Statement

*Sorry I could not get the math symbols to work properly so I did it by hand. I hope this isn't too much trouble.

Prove:
| sqrt( x ) - sqrt( y ) | <= | sqrt ( x - y ) |

for x, y >= 0

Hint: Treat the cases x >= y and x <= y separately.

I am new to proofs and we can't use calculus. It's all confusing to me and we've only just begun. The farthest we got was to the Triangle Inequality / the AGM Inequality so I assume that's the most we can do.

Homework Equations

I assume whomever is helping me already knows the AGM and Triangle Inequalities since they're the most basic of all the inequalities for proofs.

The Attempt at a Solution

I squared both sides and moved the absolute value signs to the individual variables and then in then in either case, I can remove them.

On the left side I had x + y and just removed that and -2 sqrt(x) sqrt(y) and removed that one too. On the right I had x - y so I write it as that if x > y and -y + x if x < y

Subtract x from both sides and then

y - 2 * sqrt( x ) * sqrt( y ) <= -y

I am pretty lost here. Iff x >= y then the roots would be the same and I could replace sqrt( y ) with sqrt ( x ). I would then add y to both sides and thus

2y - 2x <= 0
2(y-x) <= 0

and that's not true since x >= y and the equality would only hold true in that situation if they were equal, which we know the equality is true if y > x from substitutionPS. Thank you for taking your time to review my question and for whatever help you provide. I appreciate it very much, this question frustrated me for far too long so far.

 

[jgens]

I can't really make out what you're trying to say, but this proof follows pretty easily from the inequality:

$$0 \leq x-y \leq (\sqrt{x}+\sqrt{y})(\sqrt{x-y})$$

since $x \geq y$ (otherwise the square root of $x-y$ wouldn't exist).