TRIGONOMETRY

posted by aaron

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. Mike Oxlong

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

  2. Steve

    √(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

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