Since k,x are both positive, all the quantities are positive, and so their logarithms will have the same ordering as the quantities themselves. Given that:
Prove: (1+x)[sup]k[/sup] <= 1 + kx
is equivalent to proving:
log((1+x)[sup]k[/sup]) <= log(1 + kx)
is equivalent to proving:
k log(1+x) <= log(1 + kx)
Here’s a solution that does not use derivatives. It only depends on the fact that (1) for fixed α > 0, y[sup] α[/sup] increases (strictly) monotonically as a function of y on [1, ∞), and (2) the product of two monotonically increasing functions is monotonically increasing.
The problem is trivial if k = 1, so fix k such that 0 < k < 1. Making the substitution y = 1 + x and rearranging, the problem is equivalent to showing that if y > 1, then ky − y[sup] k[/sup] ≥ k − 1.
Observe that the right hand side of this inequality is a constant (since k is fixed) and that the inequality is satisfied with equality when y = 1. Hence, it suffices to show that ky − y[sup] k[/sup] increases monotonically as a function of y on [1, ∞).
so ky − y[sup] k[/sup] is the product of two monotonically increasing functions. Hence, ky − y[sup] k[/sup] is itself monotonically increasing, so the claim is proved.
The logarithm proof is pretty good, but there’s a variant on the derivative proof: expand the Taylor series about the origin and use the remainder formula after a few terms.