<s>[INST] 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!
 [/INST] An elementary way. Let $x=n^{1/3}\geq 1$ then it suffices to show that
$$(x^3+1)^{2/3} -x^2 <\frac{2}{3x}$$
that is
$$(x^3+1)^2<\left(x^2+\frac{2}{3x}\right)^3$$
or
$$x^6+2x^3+1<x^6+2x^3+\frac{4}{3}+\frac{8}{27x^3}$$
which trivially holds.
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