TRIGONOMETRY
posted by aaron
given real numbers x, a, b with x>= a>= b>= 0, show that sqrt x+b  sqrt xa >= sqrt x+a  sqrt xb

Mike Oxlong
sqrt x+b  sqrt xa >= sqrt x+a  sqrt xb

Steve
√(x+b)√(xa) >= √(x+a)√(xb)
since x>=a>=b, we have
x+b >= 2b
xa >= 0
x+a >= 2a >= 2b
xb >= xa
see what you can do with that
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