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

Solution 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.