Show that $(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$ for all positive integers

For $f(x) = x^{\frac{2}{3}}$, then by the Mean Value Theorem,

$(n+1)^{\frac{2}{3}}-n^{\frac{2}{3}}=\frac{f(n+1)-f(n)}{(n+1)-n} = f'(c) = \frac{2}{3}c^{-\frac{1}{3}}$ for some $c \in (n,n+1)$.

Since $f'(x)=\frac{2}{3}x^{-\frac{1}{3}}=\frac{2}{3(^3\sqrt{x})}$ is decreasing, then $c > n \implies f'(c) < f'(n) = \frac{2}{3}n^{-\frac{1}{3}}$.

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.