Elegantly Proving that $~\sqrt[5]{12}~-~\sqrt[12]5~>~\frac12$

Question

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

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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...

Answer 1

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.

Answer 2

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 $23b>16a$