- #1
asura
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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?