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$\qquad$ How could we prove, without the aid of a calculator, that $~\sqrt[5]{12}~-~\sqrt[12]5~>~\dfrac12$ ?

I have stumbled upon this mildly interesting numerical coincidence by accident, while pondering on another curios approximation, related to musical intervals. A quick computer search then also revealed that $~\sqrt[7]{12}~-~\sqrt[12]7~>~\tfrac14~$ and $~\sqrt[7]{15}~-~\sqrt[15]7~>~\tfrac13.~$ I am at a loss at finding a meaningful approach for any of the three cases. Moving the negative term to the right hand side, and then exponentiating, is —for painfully obvious reasons— unfeasible. Perhaps some clever manipulation of binomial series might show the way out of this impasse, but I fail to see how...

Michael Lugo
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Lucian
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    The gap is quite small. Maybe the only feasible way is to show that since $12>\left(\frac{4109}{2500}\right)^5$ and $5 <\left(\frac{2859}{2500}\right)^{12}$, then $\sqrt[5]{12}-\sqrt[12]{5} > \frac{4109-2859}{2500}=\frac{1}{2}$. – Giovanni Resta Mar 23 '16 at 08:45
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    I suppose you are barring approximating $\sqrt[5]{12}$ by hand( in a quite dirty fashion, but not to difficult). – S.C.B. Mar 23 '16 at 08:54
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    @GiovanniResta. If I may ask, why did you choose $2500$ ? Nice solution by the way. – Claude Leibovici Mar 23 '16 at 09:27
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    @ClaudeLeibovici Only because $\sqrt[5]{12}\approx 1.14353$ and $\sqrt[12]{5}\approx 1.64375$, so I can upper and lower bound these numbers with $1.1436$ and $1.6436$ and since they both end in "$36$", instead of fractions with denominator $10000$, like $11436/10000$ I can divide by $4$ and have fractions like $4109/2500$. – Giovanni Resta Mar 23 '16 at 09:42
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    @GiovanniResta I think that you switched $\sqrt[5]{12}$ and $\sqrt[12]{5}$ around. – Chad Shin Mar 24 '16 at 09:10
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    +1 It seems the complete set is the quadruple $15,,12,,7,,5$ such that $$\begin{aligned} \sqrt[12]{15}-\sqrt[15]{12}, &> \tfrac{1}{14}\ \sqrt[7]{15}-\sqrt[15]{7}, &> \tfrac{1}{3}\ \sqrt[5]{15}-\sqrt[15]{5}, &> \tfrac{3}{5}\ \sqrt[7]{12}-\sqrt[12]{7}, &> \tfrac{1}{4}\ \sqrt[5]{12}-\sqrt[12]{5}, &> \tfrac{1}{2}\ \sqrt[5]{7}-\sqrt[7]{5}, &> \tfrac{1}{5}\ \end{aligned}$$ with the differences being quite small. – Tito Piezas III Mar 24 '16 at 21:09

2 Answers2

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An approach using binomial series could look as follows:

For small positive $x$ and $y$ one has $$(1+x)^{1/5}>1+{x\over5}-{2x^2\over25},\qquad (1+y)^{1/12}<1+{y\over12}\ .$$ Using the ${\tt Rationalize}$ command in Mathematica one obtains, e.g., $12^{1/5}\doteq{13\over8}$. In fact $$12\cdot(8/13)^5-{18\over17}={1398\over 6\,311\,981}>0\ .$$ It follows that $$12^{1/5}>{13\over8}\left(1+{1\over17}\right)^{1/5}>{13\over8}\left(1+{1\over85}-{2\over 85^2}\right)\doteq1.64367\ .$$ In the same way Mathematica produces $5^{1/12}\doteq{8\over7}$, and one then checks that $$5\cdot (7/8)^{12}-{141\over140}=-{136\,294\,769\over2\,405\,181\,685\,760}<0\ .$$ It follows that $$5^{1/12}<{8\over7}\left(1+{1\over140}\right)^{1/12}<{8\over7}\left(1+{1\over12\cdot 140}\right)\doteq1.14354\ .$$ This solution is not as elegant as the solution found by Giovanni Resta, but the involved figures are considerably smaller.

Christian Blatter
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A quick search through the $5$th and $12$th powers of the first 25 integers yields the fact that $5 \approx \frac {8^{12}}{7^{12}}$ and that $12 \approx \frac {23^{5}}{14^{5}}$.

Thus $\sqrt[12]{5}\approx\frac8{7}$ and $\sqrt[5]{12}\approx\frac{23}{14}$.

The rest follows.

Let $\sqrt[12]{5} = \frac{16(1+a)}{14}$ and let $\sqrt[5]{12} = \frac{23(1+b)}{14}$.

Then $\sqrt[5]{12} - \sqrt[12]{5} = \frac{23(1+b)}{14} - \frac{16(1+a)}{14}= \frac{23+23b-16-16a}{14} = \frac {7}{14} + \frac{23b-16a}{14}= \frac {1}{2} + \frac{23b-16a}{14} $.

Just need to demonstrate that $23b>16a$

tomi
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    This is incorrect because $5 > 8^{12}/7^{12}$ and $12 > 23^5/14^5$. Since the inequalities go "in the same direction" you can't use them to bound the difference between the two roots. If you want to prove that $x - y \ge z$, then you need two quantities $X \le x$ and $Y \ge y$ such that $X-Y =Z \ge z$, so that the steps are $x-y \ge X - y \ge X-Y = Z \ge z$. – Giovanni Resta Mar 24 '16 at 10:02