Determine the greatest of the numbers $\sqrt2,\sqrt[3]3,\sqrt[4]4,\sqrt[5]5,\sqrt[6]6$ Determine the greatest of the numbers $$\sqrt2,\sqrt[3]3,\sqrt[4]4,\sqrt[5]5,\sqrt[6]6$$ The least common multiple of $2,3,4,5$ and $6$ is $LCM(2,3,4,5,6)=60$, so $$\sqrt2=\sqrt[60]{2^{30}}\\\sqrt[3]3=\sqrt[60]{3^{20}}\\\sqrt[4]4=\sqrt[60]{4^{15}}=\sqrt[60]{2^{30}}\\\sqrt[5]{5}=\sqrt[60]{5^{12}}\\\sqrt[6]{6}=\sqrt[60]{6^{10}}=\sqrt[60]{2^{10}\cdot3^{10}}$$ Now how do we compare $2^{30},3^{20},4^{15},5^{12}$ and $6^{10}$? I can't come up with another approach.

Let$$f(x)=x^{1/x}=e^{\log(x)/x}$$and note that $f(n)=\sqrt[n]n$, for each $n\in\Bbb N$. You have$$f'(x)=\frac{1-\log(x)}{x^2}e^{\log(x)/x},$$which is greater than $0$ on $[1,e)$ and smaller than $0$ on $(e,\infty]$. Therefore $f$ is strictly increasing on $[1,e]$ and strictly decreasing on $[e,\infty)$. So, since $e<3$ and since $3<4<5<6$,$$\sqrt[3]3>\sqrt[4]4>\sqrt[5]5>\sqrt[6]6.$$Besides, $\sqrt2=\sqrt[4]4$. And it is easy to compare $\sqrt2$ with $\sqrt[3]3$; just use the fact that $\sqrt2^6=8$ and that $\sqrt3^6=9$.
