Show that $(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$ for all positive integers 
I am trying to prove 
  $$(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$$ 
  for all positive integers.

My attempts so far have been to Taylor expand the left hand side:
$$(n+1)^{2/3} -n^{2/3}\\
=n^{2/3}\big((1+1/n)^{2/3} -1\big)\\
=n^{2/3}\left(\sum_{\alpha=0}^\infty\frac{\frac{2}{3}(\frac{2}{3}-1)\cdots(\frac{2}{3}-\alpha+1)}{\alpha!} n^\alpha-1\right)\\
=n^{2/3}\sum_{\alpha=1}^\infty\frac{\frac{2}{3}(\frac{2}{3}-1)\cdots(\frac{2}{3}-\alpha+1)}{\alpha!} n^\alpha$$
I also tried proof by induction. Assume that it's true for n=k, so that
$$(n+1)^{2/3} -n^{2/3} < \frac{2}{3}n^{-1/3}\\
n^{2/3}\big((1+1/n)^{2/3} -1\big)<\frac{2}{3} n^{-1/3}\\
(1+1/n)^{2/3} -1<\frac{2}{3} n^{-1}$$
Then I want to prove that $(n+2)^{2/3} -(n+1)^{2/3} < \frac{2}{3}(n+1)^{-1/3}$. The left hand side is:
$$(n+2)^{2/3} -(n+1)^{2/3}\\
=(n+1)^{2/3}\left[\left(1+\frac{1}{n+1}\right)^{2/3}-1^{2/3}\right]\\
<(n+1)^{2/3}\cdot \frac{2}{3} n^{-1}\\
=\frac{2}{3}\frac{(n+1)^{2/3}}{n^{-1}}$$
But this is bigger than $\frac{2}{3}(n+1)^{-1/3}$, so I am stumped!

If you are allowed to integrate,
then, for $a > 0$,
$\begin{array}\\
\int_n^{n+1} x^{-a} dx
&= \dfrac{x^{-a+1}}{-a+1}|_n^{n+1}\\
&= \dfrac{(n+1)^{-a+1}-n^{-a+1}}{-a+1}\\
\text{and}\\
\int_n^{n+1} x^{-a} dx
&\lt n^{-a}
\qquad\text{since } x^{-a} \text{ is decreasing}\\
\end{array}
$
Putting $a = \frac13$.
this becomes
$n^{-1/3}
\gt  \dfrac{(n+1)^{2/3}-n^{2/3}}{2/3}
= \dfrac32((n+1)^{2/3}-n^{2/3})
$.
Note that,
since
$\int_n^{n+1} x^{-a} dx
\gt (n+1)^{-a}
$
we get
$(n+1)^{-1/3}
\lt \dfrac32((n+1)^{2/3}-n^{2/3})
$.
