Uniform upper bound in p for difference of powers.

Question

I'm trying to prove a seemingly simple estimate, but I couldn't succeed with MVT or an induction. For $x \in [0,1]$ and $r \in \mathbb{N}$ I want to show: $$ x^r - x^{r+1} \leq \frac{1}{r+1}$$ Since I don't know how to reach the $r+1$ in the difference, I'm stuck. The uniformity will most likely come from the fact that $x \in [0,1]$ but I can't prove the inequality.

I would be very thankful and appreciative for any help!

Thanks, fixfoxi.

PS: I think $r \in\mathbb{R^+}$ would also be allowed wouldn't it?

Answer 1

You can show by differentiation that $$ f(x) = x^r - x^{r + 1} , \quad 0\leq x\leq 1 $$ has a maximum at $x_0=\frac{r}{r+1}$. At that point $$ f(x_0 ) = \frac{1}{{1 + r}}\left( {\frac{r}{{r + 1}}} \right)^r \le \frac{1}{{1 + r}}. $$

Answer 2

For fixed $r\in(0,\infty)$, consider the function $$f_r(x)=x^r-x^{r+1}=x^r(1-x)$$ defined on $[0,1]$. The first and second derivative tests will tell you that the maximum of $f_r$ occurs at$$z=\frac{r}{r+1}=1-\frac{1}{r+1}$$

Now,

$$f_r(z)=\frac{r^r}{(r+1)^{r+1}}\leq\frac{1}{r+1}$$