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

1. ## 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.

2. 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?