TRIGONOMETRY

given real numbers x, a, b with x>= a>= b>= 0, show that sqrt x+b - sqrt x-a >= sqrt x+a - sqrt x-b

1. sqrt x+b - sqrt x-a >= sqrt x+a - sqrt x-b

posted by Mike Oxlong
2. √(x+b)-√(x-a) >= √(x+a)-√(x-b)
since x>=a>=b, we have
x+b >= 2b
x-a >= 0
x+a >= 2a >= 2b
x-b >= x-a

see what you can do with that

posted by Steve

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