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$$\sum _{k=1}^n\sqrt[3]{k}\:>\frac{3}{4}n\sqrt[3]{n}$$

We have the inequality as defined as above. Clearly, one can prove it by induction; it would not be hard. However, it looks suspiciously like it could be done by Jensen's inequality; I am just not sure how. I tried the obvious substitution $f(x)=x^{1/3}$, but then this function is concave, and I cannot use it to prove the inequality. So, I was just wondering if anyone is able to use Jensen's to prove this inequality? Or is there any other inequality that anyone knows that can be used to prove this question without induction?

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  • $\begingroup$ Hint (without Jensen): Draw the function $f(x)=x^{1/3}$ and compare the sum on the left with the integral $\int_0^N f(x)dx$ $\endgroup$ – leonbloy Mar 5 at 4:58
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$$\sum_{k=1}^n\sqrt[3]{k}\geq1+\int\limits_1^n\sqrt[3]xdx=\frac{3}{4}\sqrt[3]{n^4}+\frac{1}{4}>\frac{3}{4}n\sqrt[3]n.$$

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  • $\begingroup$ how would yous how that the sum is greater than the integral though? $\endgroup$ – ONG SEE HAI HCI Mar 5 at 5:36
  • $\begingroup$ @ONG SEE HAI HCI Just draw it. $\endgroup$ – Michael Rozenberg Mar 5 at 5:37
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Probably not an answer.

Using generalized harmonic numbers, you could consider that $$\sum _{k=1}^n\sqrt[3]{k}=H_n^{\left(-\frac{1}{3}\right)}$$ and when $n$ is "sufficiently" large, $$H_n^{\left(-\frac{1}{3}\right)}=\frac{3 }{4}n^{4/3}+\frac{1}{2}n^{1/3}+\zeta \left(-\frac{1}{3}\right)+O\left(\frac{1}{n^{2/3}}\right)$$ making $$\sum _{k=1}^n\sqrt[3]{k}-\frac{3}{4}n\sqrt[3]{n}=\frac{1}{2}n^{1/3}+\zeta \left(-\frac{1}{3}\right)+O\left(\frac{1}{n^{2/3}}\right)$$ and the result is correct as soon as $n \geq 1$.

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