Given $a$ and $b$ positive integers, prove that $\min(a^{1/b},b^{1/a}) \le 3^{1/3}$

Question

The exercise is to prove that the minimum value between $a^{1/b}$ and $b^{1/a}$ is no greater than $3^{1/3}$, where $a$ and $b$ are positive integers. As it was presented in an introductory calculus class, I tried using brute Mathematics, ploting graphs, but was unable to develop and was looking for some elegant ideas.

Answer 1

Without loss of generality, let $a \leq b$ and keep $b$ be a random fixed value, then since $a, b \geq 1$ and $1/b \leq 1/a$,

$$\min{(a^{1/b}, b^{1/a})}=a^{1/b}$$

Then we clearly have maximum achieved (keeping $b$ constant) when $a=b$, or in other words, $\min{(a^{1/b}, b^{1/a})}\leq b^{1/b}$. Now the question is what value of $b$ gives us the overall maximum?

Lets consider the graph $f(x)=x^{1/x}$ on $x>0$, and see if it has any maximum points. We get (log both sides and differenciate implicitly):

$$\frac{d}{dx}\left(x^{1/x}\right)=-x^{-2+1/x}\times (\log(x)-1)$$

If we set this to $0$, then our only solution is $\log(x)=1$ giving $x=e$ (recall $x>0$).

Notice that if $x>e$, then $\frac{d}{dx}\left(x^{1/x}\right)<0$ and if $0<x<e$, $\frac{d}{dx}\left(x^{1/x}\right)>0$, hence $x=e$ is a maximum point.

Now, we can't take $b=e$ as $e$ is not an integer, but we can try $b=2$ or $b=3$ as we know $f(x)=x^{1/x}$ is strictly decreasing for $x>e$ and strictly increasing for $0<x<e$. We see that $$ 2^{1/2} < 3^{1/3} \iff 2^{6/2} < 3^{6/3} \iff 8 < 9$$

So maximum is achieved when $a=b=3$, i.e $$\min{(a^{1/b}, b^{1/a})} \leq 3^{1/3}$$

Answer 2

Suppose $a \leq b$. Then $1/a \geq 1/b$ and $a^{1/b} \leq b^{1/a}$. Seeing that $a^{1/b} \leq a^{1/a}$, It suffices to prove $a^{1/a} \leq 3^{1/3}$. Raising both sides to the $3a$-th power, it is $a^3 \leq 3^a$, or $$ a \cdots a \leq 3\cdots3, $$ where $a$ and $3$ appeared $3$ and $a$ times, respectively.

This is related to the following question: Let $N = 3a$ be an integer, we want to find a partition of $N$, which is a sequence $(a_1, a_2, \cdots, a_l)$, where $l$ is any integer, $a_i$ are all integers, and $\sum_{i=1}^l a_i = N$. We want to find $\max\{\prod_{i=1}^l{a_i}\}$.

The maximum is attained when the partition only contains $3$. We can prove this by improving the sequence to get a greater product:

• If the partition contains a 1, clearly we can merge the 1 into any term and increase the product.
• If the partition contains a $n \geq 4$, we can split $n$ into $2$ and $n-2$ and increase the product.
• Up to this step, the candidate sequence contains only 2 and 3. And 3 is more efficient than 2.

Answer 3

Hints. From (assuming) $$a<b \Rightarrow \frac{1}{a}>\frac{1}{b}$$ then (f(x)=$c^x$ is ascending for $c>1$, $g(x)=x^t$ is ascending for $\forall t>0$ for positive $x$) $$\color{red}{a}^{\frac{1}{b}}<\color{red}{a^{\frac{1}{a}}}<b^{\color{red}{\frac{1}{a}}}$$

and $$\sqrt[n+1]{n+1}<\sqrt[n]{n}, \forall n\geq3$$ leading to $$\min\left(a^{\frac{1}{b}}, b^{\frac{1}{a}}\right)=a^{\frac{1}{b}}<\color{red}{a^{\frac{1}{a}}}\leq 3^{\frac{1}{3}}, \forall a\geq3$$ and $\sqrt{2}<\sqrt[3]{3}$.