Quicker way to compare numbers without calculator

Question

Question: Find the order of $(1/2)^{1/2}$, $(1/e)^{1/e}$, $(1/3)^{1/4}$ without using calculator.

Extra constraint: You only have about 150 seconds to do it, failing to do so will eh... make you run out of time on the exam which affects the chance of admitting into graduate school!

Back to the question, when I first saw it I tried to use $(1/2.7)^{1/3}$ to approximate $(1/e)^{1/e}$, compare the $12^{th}$ power of each number and end up in getting the wrong result.

Later(too late!) I find out $f(x)=(1/x)^{1/x}$ has minimum at $x=e$ by taking the derivative $f'(x)=f(x)(ln(x)-1)/x^2$. In addition, second derivative or some kind of feeling is required to make sure it's minimum not maximum. However I feel this approach might unnecessarily take too long.

Had you seen this question, what would have been your first instinct/approach? Is there any faster way to do it?

Answer 1

I'd first raise to the power $4$, to find $1/4 < 1/3$. Or if this is a typo, raise to the power of 6 to find $1/8 < 1/9$.

For the comparisons with $1/e$, I'd take the logarithm, and do comparisons, knowing $ln(2)$ is about $0.69$, and $ln(3)$ is more than $1$.

Answer 2

$$\left(\frac13\right)^{\frac14}>\left(\frac14\right)^{\frac14}=\left(\frac12\right)^{\frac12}.$$

Knowing there is a minimum at $x=e$,

$$\left(\frac1e\right)^\frac1e<\left(\frac12\right)^{\frac12}<\left(\frac13\right)^{\frac14}.$$

Hint:

Take $\log(x^x)=x\log x$, derivative $\log x+1$, second derivative $1/x>0$: it has a minimum at $1/e$.