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

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- Apr 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)$\displaystyle ^p$<= 1+x$\displaystyle x^p$ for all x >=0 . You may assume that the derivative of $\displaystyle x^p$ is px^(p-1) if x>0.

- Apr 24th 2010, 07:59 PMmaddas
It is equivalent to show that $\displaystyle f(x) := 1+x^p - (1+x)^p$ is non-negative for $\displaystyle x\ge0$. Obviously, f(0) is non-negative. If we can show that f is increasing for $\displaystyle x\ge0$ we will be done.

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