# Inequality Challenge II

#### anemone

##### MHB POTW Director
Staff member
Prove that

$\sqrt[3]{\dfrac{2}{1}}+\sqrt[3]{\dfrac{3}{2}}+\cdots+\sqrt[3]{\dfrac{996}{995}}-\dfrac{1989}{2}<\dfrac{1}{3}+\dfrac{1}{6}+\cdots+ \dfrac{1}{8961}$

#### Opalg

##### MHB Oldtimer
Staff member
Prove that

$\sqrt[3]{\dfrac{2}{1}}+\sqrt[3]{\dfrac{3}{2}}+\cdots+\sqrt[3]{\dfrac{996}{995}}-\dfrac{1989}{2}<\dfrac{1}{3}+\dfrac{1}{6}+\cdots+ \dfrac{1}{8961}$
Not a solution, but a possible line of approach:
Start by noticing that $\dfrac{1989}2 = 995 - \dfrac12$, and $\dfrac1{8961} = \dfrac1{9\cdot995+6}.$ Next, notice that $\dfrac12 = \dfrac13+\dfrac16$. So we can subtract $\dfrac12$ from the left side, and $\dfrac13+\dfrac16$ from the right side, so that the inequality becomes $$\sqrt[3]{\dfrac{2}{1}} +\sqrt[3]{\dfrac{3}{2}} +\ldots +\sqrt[3]{\dfrac{996}{995}} - 995 < \frac19 + \frac1{12} + \frac1{15} + \ldots + \frac1{9\cdot995} + \frac1{9\cdot995+3} + \frac1{9\cdot995+6}.$$ If we write this as $$\sum_{n=1}^{995}\Bigl(\sqrt[3]{\dfrac{n+1}n} - 1\Bigr) < \sum_{n=1}^{995}\Bigl(\frac1{9n} + \frac1{9n+3} + \frac1{9n+6}\Bigr),$$ then it is tempting to think that each term on the left might be less than the corresponding term on the right, in other words $$\sqrt[3]{\dfrac{n+1}n} - 1 < \frac1{9n} + \frac1{9n+3} + \frac1{9n+6}\quad (\text{for all }n\geqslant 1).$$ I believe that is true. It certainly holds for $n=1$, because $\sqrt[3]{2} -1 \approx 0.2599$ and $\frac19 + \frac1{12} + \frac1{15} \approx 0.2611$. It also appears to hold for other values of $n$. But it is a delicate inequality – the difference between the two sides is extremely small – and I have not found any convenient way to prove it.

Edit, Oh, it's suddenly obvious!
It's just an AM-GM argument: \begin{aligned} \sqrt[3]{\dfrac{n+1}n} &= \sqrt[3]{\dfrac{3n+1}{3n} \cdot \dfrac{3n+2}{3n+1} \cdot \dfrac{3n+3}{3n+2}} \\ &< \frac13\Bigl( \dfrac{3n+1}{3n} + \dfrac{3n+2}{3n+1} + \dfrac{3n+3}{3n+2} \Bigr) \\ &= \frac13\Bigl(1 + \frac1{3n} + 1 + \frac1{3n+1} + 1 + \frac1{3n+2} \Bigr) \\ &= 1 + \frac1{9n} + \frac1{9n+3} + \frac1{9n+6}. \end{aligned}

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#### anemone

##### MHB POTW Director
Staff member
Not a solution, but a possible line of approach:
Start by noticing that $\dfrac{1989}2 = 995 - \dfrac12$, and $\dfrac1{8961} = \dfrac1{9\cdot995+6}.$ Next, notice that $\dfrac12 = \dfrac13+\dfrac16$. So we can subtract $\dfrac12$ from the left side, and $\dfrac13+\dfrac16$ from the right side, so that the inequality becomes $$\sqrt[3]{\dfrac{2}{1}} +\sqrt[3]{\dfrac{3}{2}} +\ldots +\sqrt[3]{\dfrac{996}{995}} - 995 < \frac19 + \frac1{12} + \frac1{15} + \ldots + \frac1{9\cdot995} + \frac1{9\cdot995+3} + \frac1{9\cdot995+6}.$$ If we write this as $$\sum_{n=1}^{995}\Bigl(\sqrt[3]{\dfrac{n+1}n} - 1\Bigr) < \sum_{n=1}^{995}\Bigl(\frac1{9n} + \frac1{9n+3} + \frac1{9n+6}\Bigr),$$ then it is tempting to think that each term on the left might be less than the corresponding term on the right, in other words $$\sqrt[3]{\dfrac{n+1}n} - 1 < \frac1{9n} + \frac1{9n+3} + \frac1{9n+6}\quad (\text{for all }n\geqslant 1).$$ I believe that is true. It certainly holds for $n=1$, because $\sqrt[3]{2} -1 \approx 0.2599$ and $\frac19 + \frac1{12} + \frac1{15} \approx 0.2611$. It also appears to hold for other values of $n$. But it is a delicate inequality – the difference between the two sides is extremely small – and I have not found any convenient way to prove it.
That is a good line with all excellent observations, Opalg!

Edit, Oh, it's suddenly obvious!
It's just an AM-GM argument – details to follow tomorrow.

#### Opalg

##### MHB Oldtimer
Staff member
It's just an AM-GM argument: \begin{aligned} \sqrt[3]{\dfrac{n+1}n} &= \sqrt[3]{\dfrac{3n+1}{3n} \cdot \dfrac{3n+2}{3n+1} \cdot \dfrac{3n+3}{3n+2}} \\ &< \frac13\Bigl( \dfrac{3n+1}{3n} + \dfrac{3n+2}{3n+1} + \dfrac{3n+3}{3n+2} \Bigr) \\ &= \frac13\Bigl(1 + \frac1{3n} + 1 + \frac1{3n+1} + 1 + \frac1{3n+2} \Bigr) \\ &= 1 + \frac1{9n} + \frac1{9n+3} + \frac1{9n+6}. \end{aligned}