Prove by induction (Duplicate?)

Question

Hi I am pretty sure that this is a duplicate, but I can't find it in the search:

Prove by induction:
$\frac{1}{\sqrt[\leftroot{3}3]{1}} + \frac{1}{\sqrt[\leftroot{3}3]{2}}+...+ \frac{1}{\sqrt[\leftroot{3}3]{n}}\geq n^{2/3}$

This proof is incorrect, but is it close? What am I missing?
$\frac{1}{\sqrt[\leftroot{3}3]{n}}+\frac{1}{\sqrt[\leftroot{3}3]{n+1}} = n^{2/3}+\frac{1}{\sqrt[\leftroot{3}3]{n+1}}$
Focusing on the RHS:
$n^{2/3}+\frac{1}{n+1^{1/3}}=\frac{n^{2/3}*n^{1/3}}{n^{1/3}}=\frac{1}{n^{1/3}}n$

Answer 1

We want to show that $$\color{red}{\sum_{i=1}^n \frac1{\sqrt[3]{i}}}+\frac1{\sqrt[3]{n+1}} \color{red} \ge (n+1)^{\frac23}$$

and we already know that

$$\sum_{i=1}^n \frac1{\sqrt[3]{i}}+\frac1{\sqrt[3]{n+1}}\ge n^{\frac23} + \frac1{\sqrt[3]{n+1}}$$

So your goal is to prove that

$$n^\frac23 + \frac1{\sqrt[3]{n+1}} \ge (n+1)^\frac23$$

which is equivalent to

$$\frac1{\sqrt[3]{n+1}} \ge (n+1)^\frac23- n^\frac23 $$

$$1 \ge n+1 - n^\frac23\sqrt[3]{n+1}$$

verify that the last inequality is true. Also don't forget your base case.

remark: do not omit things like summation sign or braces. Also, the last equality is wrong.

Answer 2

No induction method needed here. Just notice that \begin{align*}&\frac{1}{\sqrt[3]{1}}+\frac{1}{\sqrt[3]{2}}+\frac{1}{\sqrt[3]{3}}+\cdots+\frac{1}{\sqrt[3]{n}}\\\geq & \frac{1}{\sqrt[3]{n}}+\frac{1}{\sqrt[3]{n}}+\frac{1}{\sqrt[3]{n}}+\cdots+\frac{1}{\sqrt[3]{n}}\\=& \frac{n}{\sqrt[3]{n}}\\= & n^{\frac{2}{3}}\end{align*}