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Prove that for any integers $m$ and $n$ greater than $1$, at least one of the numbers $\sqrt[n]{m}$ or $\sqrt[m]{n}$ is not greater than $\sqrt[3]{3}$.

My attempt goes something along the lines of stating that for $(n,m)\in\{(2,2),(2,3),(3,2),(3,3)\}$ is obvious, and for $n,m\geq4$, if $n=m$ then one needs to prove $\log_ 3n<n/3$, and if $n\ne m$, I have no idea.

As for proof of $\log_ 3n<n/3$, is it enough to state that the equality holds for $n=3$ and that after that, for $n>3$, the slope of $f(n)=\log_3n$ is smaller than $1/3$ (yields from $f'(n)=(n\ln3)^{-1}<3^{-1}$ because $\ln3>1$), so the function $f(n)$ doesn't raise fast enough to catch up to $n/3$, so there will be no more intersections?

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Are you sure you don't want to re-tag this? Surely, it involves calculus? Perhaps add [elementary]-and/or-number theory? BTW: Nice questions +1 ! – amWhy Feb 7 '13 at 1:23
    
You don't need the numbers to be greater than $1$... it reads nicely if they're just "natural numbers". – mjqxxxx Feb 7 '13 at 15:17
    
@amWhy, the accepted answer proves that this problem doesn't require calculus at all. Even logarithms are not needed – You're In My Eye Jul 1 at 16:30
up vote 3 down vote accepted

You want to show that $$ \min\left({\sqrt[m]{n}, \sqrt[n]{m}}\right) \le \sqrt[3]{3}. $$ Raising both sides to the $(3mn)$-th power, we see that this is equivalent to $$ \min\left(m^{3m}, n^{3n}\right)\le 3^{mn}. $$ Assume with no loss of generality that $m \le n$; then $\min\left(m^{3m},n^{3n}\right)=m^{3m}$ and $3^{mn}\ge 3^{m^2}$. Because $ m^3 \le 3^m $ for all $m\ge 1$ (*), we have $$ m^{3m} =\left[m^3\right]^{m}\le \left[3^{m}\right]^{m} = 3^{m^2} \le 3^{mn}$$ and we're done.


(*) The proof that $m^3 \le 3^m$ for all $m \ge 1$ is by induction: it holds for $m=1,2,3$ by direct inspection, and in going from $m$ to $m+1$ for $m\ge 3$, the right-hand side is multiplied by $3$, while the left-hand side is multiplied by $$\frac{(m+1)^3}{m^3}=1+\frac{3}{m}+\frac{3}{m^2}+\frac{1}{m^3} \le 1+\frac{3}{3}+\frac{3}{3^2}+\frac{1}{3^3}<3.$$

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The induction proof is a very good way, since the tag is 'algebra-precalculus', so using calculus is forbidden. +1 for you – You're In My Eye Jul 1 at 16:29

Strictly speaking, it fails for $m=n=3$ unless you make your $\lt$ into $\le$. If you prove it for $n=m$, you can argue that for $n \gt m, \sqrt[n]m \lt \sqrt[m]m$, so you only need to worry about equality. Using calculus, you can show that $\sqrt[x]x$ is decreasing for $x \gt 3$ and you are there.

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