Inequality with cube roots $$(\sqrt{n}+1)^{1/3}-(\sqrt{n}-2)^{1/3} \geq (\sqrt{n+1}+1)^{1/3}-(\sqrt{n+1}-2)^{1/3}$$
$$n\in \mathbb{N}$$
I come upon this inequality when trying to use prove series declension for the Leibnit'z criterion.
I haven't been able to find a good solution to this.
 A: Hint: Consider the function :$f(x)=(\sqrt{x}+1)^{\frac{1}{3}}-(\sqrt{x}-2)^{\frac{1}{3}}$ ,$ \forall x >4$.
$f'(x)=\frac{1}{6\sqrt{x}}(\frac{1}{\sqrt[3]{(\sqrt{x}+1)^2}} -\frac{1}{\sqrt[3]{(\sqrt{x}-2)^2}})=\frac{1}{6\sqrt{x}}\frac{\sqrt[3]{(\sqrt{x}-2)^2}-\sqrt[3]{(\sqrt{x}+1)^2}}{\sqrt[3]{(\sqrt{x}+1)^2}\times\sqrt[3]{(\sqrt{x}-2)^2}}<0 $ ,$\forall x >4$.
which gives us that $f(x)$ is monotonically decreasing.
so for $n<n+1$,we have $f(n)>f(n+1)$
hence,$$(\sqrt{n}+1)^{1/3}-(\sqrt{n}-2)^{1/3} \geq (\sqrt{n+1}+1)^{1/3}-(\sqrt{n+1}-2)^{1/3}$$
$$n\in \mathbb{N}$$
A: We have $$a-b=\frac{a^3-b^3}{a^2+ab+b^2}.$$
From this we obtain
$$x^{1/3}-y^{1/3}=\frac{x-y}{x^{2/3}+(xy)^{1/3}+y^{2/3}}.$$
Using this we get
$$(\sqrt{n}+1)^{1/3}-(\sqrt{n}-2)^{1/3} = \frac{(\sqrt{n}+1)-(\sqrt{n}-2)}{(\sqrt{n}+1)^{2/3}+(\sqrt{n}+1)^{1/3}(\sqrt{n}-2)^{1/3}+(\sqrt{n}-2)^2/3}=
\frac{3}{(\sqrt{n}+1)^{2/3}+(\sqrt{n}+1)^{1/3}(\sqrt{n}-2)^{1/3}+(\sqrt{n}-2)^2/3}.$$
Since the numerator (as a function of $n$) is increasing, the whole thing is decreasing.
(To be more precise, it can be seen that the numerator is increasing if $\sqrt n-2 \ge 0$, i.e., if $n\ge4$.)
