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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!

Robert Z
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supercoolphysicist
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    Write the required inequality in the form $$\left(1+\frac{1}{n}\right)^{2/3}<1+\frac{2}{3n}.$$ The above ineq is true by Bernoulli's ineq. You can also prove using Taylor's thm: $(1+x)^{2/3}=1+\frac{2}{3}x+R(x)$, and show that $R(x)<0$. – Batominovski Mar 24 '20 at 18:37
  • The standard Bernoulli's inequality is for integer exponents only. This, cubing, becomes $(1+1/n)^2 < (1+2/(3n))^3$ or $1+2/n+1/n^2 < 1+2/n+4/(3n^2)+8/(27n^3)$ or $1/n^2 < 4/(3n^2)+8/(27n^3)$ which is true, barely. – marty cohen Mar 24 '20 at 20:49
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    @martycohen Bernoulli's Inequality has a version for every exponent, not just integer exponents. Yes, the standard version is for integer exponents, but the version for every real exponent is also widely cited as just Bernoulli's Inequality. And it is not any less true if the generalized version is used. – Batominovski Mar 25 '20 at 21:28

5 Answers5

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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 $c > n \implies f'(c) < f'(n) = \frac{2}{3}n^{-\frac{1}{3}}$.

bloomers
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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.

Robert Z
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Try multiplying both sides by $n^{1/3}$, expanding $(n(n+1)^2)^{1/3}$ to $(n^3+2n^2+n)^{1/3}$, and moving $n$ from the left hand side to the right hand side, which rewrites the inequality as

$$(n^3+2n^2+n)^{1/3}\lt n+{2\over3}$$

Cubing both sides (which is OK since $x^3\lt y^3\iff x\lt y$) turns the inequality to prove into

$$n^3+2n^2+n\lt n^3+2n^2+{4\over3}n+{8\over27}$$

which is obviously true.

Barry Cipra
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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}) $.

marty cohen
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It seems nobody can resist this one :) $$\begin{align}(n+1)^{2/3}-n^{2/3}&=\left((n+1)^{1/3}-n^{1/3}\right)\left((n+1)^{1/3}+n^{1/3}\right)\\ &=\frac{\left(n+1-n\right)\left((n+1)^{1/3}+n^{1/3}\right)}{(n+1)^{2/3}+n^{1/3}(n+1)^{1/3}+n^{2/3}}\\ &=\frac1{n^{1/3}\left(\frac{(n+1)^{2/3}}{n^{1/3}(n+1)^{1/3}+n^{2/3}}+1\right)}\\ &=\frac1{n^{1/3}\left(\frac{\left(\frac{n+1}{n}\right)^{1/3}}{1+\left(\frac{n}{n+1}\right)^{1/3}}+1\right)}\end{align}$$ And there you have it: $\frac n{n+1}<1$, so $1+\left(\frac n{n+1}\right)^{1/3}<2$ and $$\frac{\left(\frac{n+1}{n}\right)^{1/3}}{1+\left(\frac{n}{n+1}\right)^{1/3}}+1>\frac{\left(\frac{n+1}n\right)^{1/3}}2+1>\frac12+1=\frac32$$ So $$(n+1)^{2/3}-n^{2/3}<\frac1{\frac32n^{1/3}}$$

user5713492
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