Elegantly Proving that $~\sqrt[5]{12}~-~\sqrt[12]5~>~\frac12$ $\qquad$ How could we prove, without the aid of a calculator, that $~\sqrt[5]{12}~-~\sqrt[12]5~&gt;~\dfrac12$ ?




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 $~\sqrt[7]{12}~-~\sqrt[12]7~&gt;~\tfrac14~$ and $~\sqrt[7]{15}~-~\sqrt[15]7~&gt;~\tfrac13.~$ 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 &mdash;for painfully obvious reasons&mdash; unfeasible. Perhaps some clever manipulation of binomial series might show the way out of this impasse, but I fail to see how...
An approach using binomial series could look as follows:

For small positive $x$ and $y$ one has
$$(1+x)^{1/5}&gt;1+{x\over5}-{2x^2\over25},\qquad (1+y)^{1/12}&lt;1+{y\over12}\ .$$
Using the ${\tt Rationalize}$ command in Mathematica one obtains, e.g., $12^{1/5}\doteq{13\over8}$. In fact
$$12\cdot(8/13)^5-{18\over17}={1398\over 6\,311\,981}&gt;0\ .$$
It follows that
$$12^{1/5}&gt;{13\over8}\left(1+{1\over17}\right)^{1/5}&gt;{13\over8}\left(1+{1\over85}-{2\over 85^2}\right)\doteq1.64367\ .$$
In the same way Mathematica produces $5^{1/12}\doteq{8\over7}$, and one then checks that
$$5\cdot (7/8)^{12}-{141\over140}=-{136\,294\,769\over2\,405\,181\,685\,760}&lt;0\ .$$
It follows that
$$5^{1/12}&lt;{8\over7}\left(1+{1\over140}\right)^{1/12}&lt;{8\over7}\left(1+{1\over12\cdot 140}\right)\doteq1.14354\ .$$
This solution is not as elegant as the solution found by Giovanni Resta, but the involved figures are considerably smaller.
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 $23b&gt;16a$