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Math Help - show that (1+x)^p <= 1+x^p

  1. #1
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    show that (1+x)^p <= 1+x^p

    Let p be a fixed real number satisfying 0<p<=1 show that (1+x) ^p<= 1+x x^p for all x >=0 . You may assume that the derivative of x^p is px^(p-1) if x>0.
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  2. #2
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    It is equivalent to show that f(x) := 1+x^p - (1+x)^p is non-negative for x\ge0. Obviously, f(0) is non-negative. If we can show that f is increasing for x\ge0 we will be done.

    Differentiate to get f'(x) = px^{p-1} - p(1+x)^{p-1} = p(\frac1{x^{1-p}} - \frac1{(1+x)^{1-p}}). Can you tell that this is non-negative?
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