TITLE: Can we minimize $3^m 2^{n/m}$, given $n$?
QUESTION [5 upvotes]: If we are given $n$, a positive real, can we find a the positive real $m$ that minimizes the function:
$$3^m 2^{n/m}$$
I'd prefer to find the function that gives a value for $m$, but I'm also interested in asymptotic bounds for $m$.

REPLY [2 votes]: Minimizing $3^m2^{n/m}$ is the same as minimizing
$$\log 3^m2^{n/m} = m\log 3 + \frac nm\log2,$$
which we can solve by setting the derivative in $m$ equal to $0$.

REPLY [2 votes]: If you take the logarithm, your expression turns into
$$m\ln 3+\frac{n\ln 2}{m}.$$
By the arithmetic-geometric-mean inequality
$$ m\ln 3+\frac{n\ln 2}{m}\ge 2\sqrt{n\ln 3\ln2}$$
with equality iff $m\ln 3=\frac{n\ln 2}{m}$, that is iff $m=\sqrt{n\log_32}$.