Proving: sqrt(x+b) - sqrt(x-a) >= sqrt(x+a) - sqrt(x-b)

[asura]

Homework Statement

show that sqrt(x+b) - sqrt(x-a) >= sqrt(x+a) - sqrt(x-b)
where x >= a >= b >= 0

Homework Equations

none

The Attempt at a Solution

My instructor said that we had to use an indirect proof.
The give statement is "if x >= a >= b >= 0, then sqrt(x+b) - sqrt(x-a) >= sqrt(x+a) - sqrt(x-b)"
So for the proof, the statement is "if x >= a >= b >= 0, then sqrt(x+b) - sqrt(x-a) < sqrt(x+a) - sqrt(x-b)"

By manipulating sqrt(x+b) - sqrt(x-a) < sqrt(x+a) - sqrt(x-b), I get abs(a) < abs(b).

Heres the algebra
sqrt(x+b) - sqrt(x-a) < sqrt(x+a) - sqrt(x-b)
sqrt(x+b) + sqrt(x-b) < sqrt(x+a) + sqrt(x-a)
x + b + 2 sqrt(x^2 - b^2) + x - b < x + a + 2 sqrt(x^2 - a^2) + x - a
2x + 2 sqrt(x^2 - b^2) < 2x + 2 sqrt(x^2 - a^2)
sqrt(x^2 - b^2) < sqrt(x^2 - a^2)
x^2 - b^2 < x^2 - a^2
a^2 < b^2
abs(a) < abs(b)

abs(a) < abs(b) contradicts the given: x >= a >= b >= 0
So my question is does that automatically make "if x >= a >= b >= 0, then sqrt(x+b) - sqrt(x-a) >= sqrt(x+a) - sqrt(x-b)" true? or did I miss something?

 

[HallsofIvy]

If assuming sqrt(x+b) - sqrt(x-a) < sqrt(x+a) - sqrt(x-b), leads to a contradictionthen it can't be true. If it can't be true what is true?

 

[tiny-tim]

sqrt(x+b) - sqrt(x-a) < sqrt(x+a) - sqrt(x-b)
sqrt(x+b) + sqrt(x-b) < sqrt(x+a) + sqrt(x-a)
x + b + 2 sqrt(x^2 - b^2) + x - b < x + a + 2 sqrt(x^2 - a^2) + x - a
2x + 2 sqrt(x^2 - b^2) < 2x + 2 sqrt(x^2 - a^2)
sqrt(x^2 - b^2) < sqrt(x^2 - a^2)
x^2 - b^2 < x^2 - a^2
a^2 < b^2
abs(a) < abs(b)

Hi asura!

(feel free to copy √ and ² and ≤, and anything else you like, for future use )

Personally, I think "inequations" (with < or ≤) are really confusing.

Hint: it's often clearer to replace the < by a minus, so you get:

[√(x+b) + √(x-b)]² - [√(x+a) + √(x-a)]²
= …
= 2√(x² - b²) - 2√(x² - a²),
and carry on from there (starting with "But b² ≤ a²", so …).