Let p be a fixed real number satisfying 0<p<=1 show that (1+x) <= 1+x for all x >=0 . You may assume that the derivative of is px^(p-1) if x>0.

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- April 24th 2010, 07:38 PMtn11631show that (1+x)^p <= 1+x^p
Let p be a fixed real number satisfying 0<p<=1 show that (1+x) <= 1+x for all x >=0 . You may assume that the derivative of is px^(p-1) if x>0.

- April 24th 2010, 07:59 PMmaddas
It is equivalent to show that is non-negative for . Obviously, f(0) is non-negative. If we can show that f is increasing for we will be done.

Differentiate to get . Can you tell that this is non-negative?