, 2015

Solution:
Let $S$ be the set of all positive integers $n$ such that
$$\phi(n) \cdot \tau(n) \geq \sqrt{\frac{n^{3}}{3}}$$
Define a function $\Phi$ on all positive integers $n$ by
$$\Phi(n)=\frac{\phi(n)^{2} \cdot \tau(n)^{2}}{n^{3}}$$
An important observation is that $\Phi$ has the property that for every relatively prime positive integers $m, n$, we have $\Phi(m n)=\Phi(m) \Phi(n)$.
Define another function $\psi$ on all ordered pairs $(a, p)$ of positive integer $a$ and prime number $p$ as follows:
$$\psi(a, p):=\frac{(a+1)^{2}(1-1 / p)^{2}}{p^{a}}$$
If we express $n$ in its canonical form as $n=\prod_{i=1}^{k} p_{i}^{a_{i}}$, then we have
$$\Phi(n)=\prod_{i=1}^{k} \psi\left(a_{i}, p_{i}\right)$$
Therefore, $S$ is actually the set of all $n=\prod_{i=1}^{k} p_{i}^{a_{i}}$ such that
$$\prod_{i=1}^{k} \psi\left(a_{i}, p_{i}\right) \geq \frac{1}{3}$$
It is straightforward to establish the following: for every prime $p$ and positive integer $a$,
- if $p \geq 11$, then $\psi(a, p)<\frac{1}{3}$;
- if $p=5$ and $a \geq 2$, then $\psi(a, p)<\frac{1}{3}$;
- if $p=3$ and $a \geq 3$, then $\psi(a, p)<\frac{1}{3}$;
- if $p=2$ and $a \geq 5$, then $\psi(a, p)<\frac{1}{3}$;
- $\psi(a, p)$ is always less than $\frac{1}{\sqrt{3}}$ unless $(a, p)=(1,3)$ where $\psi(1,3)=\frac{16}{27}$.
The data above shows that in the case $n=p^{a}$, in which there is only one prime dividing $n$, there are at most 8 possible $n$ in $S$: $2^{1}, 2^{2}, 2^{3}, 2^{4}, 3^{1}, 3^{2}, 5^{1}, 7^{1}$. If $n$ is divisible by at least two distinct primes,
then one of them must be $3^{1}$ and $3^{1}$ fully divides $n$ (that is, $3^{2} \nmid n$). In the latter case $n=3 \cdot n_{0}$. Write $n_{0}=\prod_{j=1}^{l} q_{j}^{b_{j}}$. In order for $\Phi(n) \geq \frac{1}{3}$, we require
$$\prod_{j=1}^{l} \psi\left(b_{j}, q_{j}\right) \geq \frac{9}{16}$$
This is only possible when $(b, q)=(2,2)$, where $n=12$. Also, note that when $n=1, \Phi(1)=1$, so $1 \in S$. Hence, there are at most 10 (a finite number of) possible values of $n$ in $S$.