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let $n$ be postive integer numbers,and such $n>2$,show that

$$ 8\,\,\left.\right\vert\,\,\left\lfloor% \left(\vphantom{\Large A}\sqrt[3]{n\,} + \sqrt[3]{n + 2\,}\,\right)^{3}% \right\rfloor + 1 $$ where $\left\lfloor x\right\rfloor$ is is the largest integer not greater than $x$,

My try: since

$$(a+b)^3=a^3+b^3+3ab(a+b)$$ then $$(\sqrt[3]{n}+\sqrt[3]{n+2})^3=n+n+2+3\sqrt[3]{n^2+2n}(\sqrt[3]{n}+\sqrt[3]{n+2})$$ so $$[(\sqrt[3]{n}+\sqrt[3]{n+2})^3]=2n+2+[3\sqrt[3]{n^2+2n}(\sqrt[3]{n}+\sqrt[3]{n+2})]$$

then I can't,Thank you

math110
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2 Answers2

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Claim For $n >2$ we have

$$\left\lfloor% \left(\vphantom{\Large A}\sqrt[3]{n\,} + \sqrt[3]{n + 2\,}\,\right)^{3}% \right\rfloor =8n+7$$

This is equivalent to

$$8n+7 \leq \left(\vphantom{\Large A}\sqrt[3]{n\,} + \sqrt[3]{n + 2\,}\,\right)^{3}< 8n+8 \Leftrightarrow \\ 8n+7 \leq 2n+2+[3\sqrt[3]{n^2+2n}(\sqrt[3]{n}+\sqrt[3]{n+2})] < 8n+8 \Leftrightarrow \\ 6n+5 \leq3\sqrt[3]{n^2+2n}(\sqrt[3]{n}+\sqrt[3]{n+2}) < 6n+6 $$

Hopefully this part is right: By Jensen

$$\sqrt[3]{n}+\sqrt[3]{n+2}< 2 \sqrt[3]{\frac{n+n+2}{2}}=2\sqrt[3]{n+1}$$ also $$3\sqrt[3]{n^2+2n}<3\sqrt[3]{n^2+2n+1}=3\sqrt[3]{(n+1)^2}\,.$$

Multiplying these two inequalities yields the upperbound.

For the lowerbound we know

$$3\sqrt[3]{n^2+2n}(\sqrt[3]{n}+\sqrt[3]{n+2}) =3\sqrt[3]{n^3+2n^2}+3\sqrt[3]{n^3+4n^2+4n}$$

Now, I think you can prove:

$$3\sqrt[3]{n^3+4n^2+4n} \geq 3n+3.5$$ and for $n > 2$ we have $$3\sqrt[3]{n^3+2n^2} > 3n+1.5$$

Much Simpler Solution

By AM-GM

$$\left( \sqrt[3]{n\,} + \sqrt[3]{n + 2\,} \right)^3 > 8 \sqrt{n^2+2n\,} >8n+7$$

with the last inequality being true for $n >3$. The case $n=3$ is trivial to check.

As Calvin pointed, by Jensen we have

$$\left(\vphantom{\Large A}\sqrt[3]{n\,} + \sqrt[3]{n + 2\,}\,\right)^{3} < \left(2\vphantom{\Large A}\sqrt[3]{n+1\,} \right)^{3} =8n+8$$

N. S.
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    For the upper bound, you could have done it directly (before expanding) by Jensen's, since $ \sqrt[3]{n} + \sqrt[3]{n+2} \leq 2 \sqrt[3] {n+1} $ – Calvin Lin Nov 15 '13 at 05:34
  • P.S. I am not sure if 3.5 and 1.5 are the best choices for the last part, 3 and 2 almost work , is just that 2 is just a tad to big but 3 can be increased..... – N. S. Nov 15 '13 at 05:35
  • @CalvinLin Woow I missed the obvious, should be going to sleep. I typed at this post for 20 minutes, trying to get the right bounds :) – N. S. Nov 15 '13 at 05:36
  • @CalvinLin And I also missed that the lowerbound follows from AM-GM – N. S. Nov 15 '13 at 05:42
  • +1 Nice. Not exactly AM-GM, but it's the power-mean inequalities. – Calvin Lin Nov 15 '13 at 05:49
  • @CalvinLin AM-GM: $\sqrt[3]{n,} + \sqrt[3]{n + 2,} > 2 \sqrt{\sqrt[3]{n,} \sqrt[3]{n + 2,}}$ and cube it :) – N. S. Nov 15 '13 at 05:52
  • ahhahaha. I did it as $ (\frac{ \sqrt[3]{n} + \sqrt[3]{n+2} } { 2} )^3 \geq \sqrt{n(n+2)}$, so it was the 1/3 power vs the 0 power lol. You should rewrite your solution. – Calvin Lin Nov 15 '13 at 05:54
  • Nice solution,Thank you +1 – math110 Nov 15 '13 at 06:44
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This solution proves an even somewhat stronger statement. First, have

$$S_1 = (\sqrt[3]{n + 2} + \sqrt[3]{n})^3 = (n + 2) + 3(\sqrt[3]{n + 2})^2\sqrt[3]{n} + 3\sqrt[3]{n + 2}\,(\sqrt[3]{n})^2 + n \tag{1}\label{eq1A}$$

$$S_2 = (\sqrt[3]{n + 2} - \sqrt[3]{n})^3 = (n + 2) - 3(\sqrt[3]{n + 2})^2\sqrt[3]{n} + 3\sqrt[3]{n + 2}\,(\sqrt[3]{n})^2 - n \tag{2}\label{eq2A}$$

Adding these values gives

$$S_1 + S_2 = 2n + 4 + 6\sqrt[3]{n + 2}\,(\sqrt[3]{n})^2 \tag{3}\label{eq3A}$$

Letting $k_1 = \sqrt[3]{n + 2}\,(\sqrt[3]{n})^2$, then

$$\left(n + \frac{2}{3}\right)^3 - k_1^3 = \left(n^3 + 2n^2 + \frac{4n}{3} + \frac{8}{27}\right) - (n^3 + 2n^2) = \frac{4n}{3} + \frac{8}{27} \gt 0 \tag{4}\label{eq4A}$$

Thus, $k_1 \lt n + \frac{2}{3}$ so, from \eqref{eq3A}, we get

$$S_1 + S_2 \lt 2n + 4 + 6\left(n + \frac{2}{3}\right) = 8n + 8 \tag{5}\label{eq5A}$$

From \eqref{eq1A} and \eqref{eq2A}, we also have

$$S_1 - S_2 = 2n + 6(\sqrt[3]{n + 2})^2\sqrt[3]{n} \tag{6}\label{eq6A}$$

Using $k_2 = (\sqrt[3]{n + 2})^2\sqrt[3]{n}$, then

$$k_2^3 - \left(n + \frac{7}{6}\right)^3 = (n^3 + 4n^2 + 4n) - \left(n^3 + \frac{7n^2}{2} + \frac{49n}{12} + \frac{343}{216}\right) = \frac{n^2}{2} - \frac{n}{12} - \frac{343}{216} \tag{7}\label{eq7A}$$

Let $f(n) = \frac{n^2}{2} - \frac{n}{12} - \frac{343}{216}$. With $f(2) \approx 0.245$ and $f'(n) = n - \frac{1}{12} \gt 0$, then $f(n) \gt 0$ for all $n \ge 2$. Thus, $k_2 \gt n + \frac{7}{6}$ so, from \eqref{eq6A}, then

$$S_1 - S_2 \gt 2n + 6\left(n + \frac{7}{6}\right) = 8n + 7 \tag{8}\label{eq8A}$$

Since $S_2 \gt 0$, we get from \eqref{eq1A}, \eqref{eq5A} and \eqref{eq8A} that

$$8n + 7 \lt S_1 \lt 8n + 8 \;\;\;\to\;\;\; 8n + 7 \lt (\sqrt[3]{n} + \sqrt[3]{n + 2})^3 \lt 8n + 8 \tag{9}\label{eq9A}$$

Thus, $\lfloor(\sqrt[3]{n} + \sqrt[3]{n + 2})^3\rfloor + 1 \equiv 0 \pmod{8}$ for all integers $n \ge 2$.

John Omielan
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