How can this be proven? (maths)

The answer is to use logarithms.

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)

Now since 0 < k <= 1, and x > 0

k log(1+x) <= log(1+x)
(1+x) <= (1+kx)

and hence:

k log(1+x) <= log(1+x) <= log(1+kx)

and you’re done.

If k is in (0, 1], then 1 + x > 1 + kx.

oops. Sigh. My first thought was derivatives too. Should have stuck with it.

If k is restricted to integers, this is very easily proved by induction on k …

I think you misread the statement. If k is an integer greater than 1, the inequality is false (it’s reversed).

Well, in this case, “k restricted to integers” means k = 1, and that case is very easy to prove by induction ;).

You’re right, I did. But I successfully proved the reversed inequality for k>=1. :slight_smile:

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
k yy[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 k yy[sup] k[/sup] increases monotonically as a function of y on [1, ∞).

Indeed, we have that

k yy[sup] k[/sup] = (y[sup]1 − k[/sup] − k[sup] −1[/sup]) k y[sup] k[/sup],

so k yy[sup] k[/sup] is the product of two monotonically increasing functions. Hence, k yy[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.